Compute the area of each triangle. Round to the nearest tenth.

Compute The Area Of Each Triangle. Round To The Nearest Tenth.

Answers

Answer 1

The triangle ΔDEF has the following coordinates

[tex]\lbrace D(-1,6),E(-4,-6),F(3,-5)\rbrace[/tex]

To find the area of a triangle in coordinate geometry, we have a formula. Given 3 vertices A(x1, y1), B(x2,y2) and C(x3,y3), the area of this triangle is given by

[tex]Area(\Delta ABC)=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|[/tex]

Using this formula for our problem, we have

[tex]Area_{\Delta DEF}=\frac{1}{2}|(-1)((-6)-(-5))+(-4)((-5)-6)+3(6_{}-(-6))|[/tex]

Solving this equation, we have

[tex]\begin{gathered} Area_{\Delta DEF}=\frac{1}{2}|(-1)((-6)-(-5))+(-4)((-5)-6)+3(6_{}-(-6))| \\ =\frac{1}{2}|(-1)((-6+5)+(-4)(-5-6)+3(6_{}+6)| \\ =\frac{1}{2}|(-1)(-1)+(-4)(-11)+3(12)| \\ =\frac{1}{2}|1+44+36| \\ =\frac{1}{2}|81| \\ =\frac{81}{2} \\ =40.5 \end{gathered}[/tex]

And this is our answer Area(ΔDEF) = 40.5


Related Questions

Two legs of a step ladder are each 4 metres long. The angle formed between the two legs is 30degrees.Make a labelled scale drawing of the ladder using the scale Icm=0.5 metres and fill in the blanksbelow.

Answers

assume the figure as two step ladder

I need help to solve by using the information provided to write the equation of each circle! Thanks

Answers

Explanation

For the first question

We are asked to write the equation of the circle given that

[tex]\begin{gathered} center:(13,-13) \\ Radius:4 \end{gathered}[/tex]

The equation of a circle is of the form

[tex](x-a)^2+(y-b)^2=r^2[/tex]

In our case

[tex]\begin{gathered} a=13 \\ b=-13 \\ r=4 \end{gathered}[/tex]

Substituting the values

[tex](x-13)^2+(y+13)^2=4^2[/tex]

For the second question

Given that

[tex](18,-13)\text{ and \lparen4,-3\rparen}[/tex]

We will have to get the midpoints (center) first

[tex]\frac{18+4}{2},\frac{-13-3}{2}=\frac{22}{2},\frac{-16}{2}=(11,-8)[/tex]

Next, we will find the radius

Using the points (4,-3) and (11,-8)

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put the numbers in order from least to greatest2.3,12/5,5/2,2.2,21/10

Answers

Express the fraction in terms of decimal.

[tex]\frac{12}{5}=2.4[/tex][tex]\frac{5}{2}=2.5[/tex][tex]\frac{21}{10}=2.1[/tex]

The numbers are,

2.3, 2.4, 2.5, 2.2, 2.1.

Now we arrange the number from least to greatest.

[tex]2.1,2.2,2.3,2.4,2.5[/tex]

So answer is,

[tex]\frac{21}{10},2.2,2.3,\frac{12}{5},\frac{5}{2}[/tex]

Hello this is a multi step question and I am struggling to help my son with this. It is 1 of 3 so hoping to get guidance with this first one to be able to know how to apply it to the others in his activities. Thank you as I know this is multiple steps and time consuming. The help is greatly appreciated as a parent.

Answers

In the first part of this problem, we must compute some statistic variables of two distributions:

0. the mean value,

,

1. the median,

,

2. the standard deviation.

,

3. the interquartile range.

1. The mean of a data set is the sum of all the data divided by the count n:

[tex]\mu=\frac{x_1+x_2+\cdots+x_n}{n}\text{.}[/tex]

2. The median is the data value separating the upper half of a data set from the lower half, it is computed following these steps:

• arrange data values from lowest to the highest value,

,

• the median is the data value in the middle of the set

,

• if there are 2 data values in the middle the median is the mean of those 2 values.

3. The standard deviation for a sample data set is given by the following formula:

[tex]\sigma=\sqrt[]{\frac{(x_1-\mu)^2+(_{}x_2-\mu)^2+\cdots+(x_n-\mu)^2}{n-1}_{}}\text{.}[/tex]

4. The interquartile range (IQR) is given by:

[tex]\text{IQR}=Q_3-Q_1\text{.}[/tex]

Where Q_1 and Q_3 are the first and third quartiles. The lowest quartile (Q1) covers the smallest quarter of values in your dataset.

--------------

Using the definitions above, we compute the mean, the median and the standard deviation for the samples taken by Manuel and Gretchen.

Manuel's sample

• Sample = {3, 6, 8, 11, 12, 8, 6, 3, 10, 5, 14, 9, 7, 10, 8}

,

• Count = 15

1. Mean

Using the formula above, we get:

[tex]\mu=\frac{120}{5}=8.[/tex]

2. Median

We order the data set:

[tex]3,3,5,6,6,7,8,(8),8,9,10,10,11,12,14.[/tex]

From the ordered data set, we see that the central number 8 divides the data set into two equal parts.

So the median of this sample is:

[tex]\bar{x}=8.[/tex]

3. Standard deviation

Using the formula above, we get:

[tex]\sigma=\sqrt[]{\frac{138}{15-1}}\cong3.14.[/tex]

4. Interquartile range

Dividing the data sample into quartiles, we have:

[tex]3,3,5,6|6,7,8|8|8,9,10|10,11,12,14.[/tex]

We have:

• Q_1 = 6,

,

• Q_3 = 10.

So the interquartile range is:

[tex]\text{IQR }=Q_3-Q_1=10-6=4.[/tex]

Gretchen's sample

• Sample = {22, 4, 7, 8, 12, 15, 10, 7, 9, 6, 13, 3, 8, 10, 10}

,

• Count = 15

1. Mean

[tex]\mu=\frac{144}{15}=9.6.[/tex]

2. Median

We order the data set:

[tex]3,4,6,7,7,8,8,(9),10,10,10,12,13,15,22.[/tex]

From the ordered data set, we see that the central number 8 divides the data set into two equal parts.

So the median of this sample is:

[tex]\bar{x}=9.[/tex]

3. Standard deviation

[tex]\sigma=\sqrt[]{\frac{307.6}{15-1}}\cong4.69.[/tex]

4. Interquartile range

Dividing the data sample into quartiles, we have:

[tex]3,4,6,7|7,8,8|9|10,10,10|12,13,15,22.[/tex]

We have:

• Q_1 = 7,

,

• Q_3 = 12.

So the interquartile range is:

[tex]\text{IQR }=Q_3-Q_1=12-7=5.[/tex]

Answers

Manuel's sample

0. Mean = 8

,

1. Median = 8

,

2. Standard deviation ≅ 3.14

,

3. Interquartile range = 4

Gretchen's sample

0. Mean = 9.6

,

1. Median = 9

,

2. Standard deviation ≅ 4.69

,

3. Interquartile range = 5

A retail clothing store offers customers an opportunity to open up a credit card during checkout. One location of the retail clothing store states that the number of credit cards, A, that are opened t months since January can be modeled by the function A(t) = 15 + 3t. The number of credit cards opened at another location, B, is defined by the function B(t) = 25 − t. What is an expression that can be used to determine the total amount of credit cards opened at the two locations?

(A + B)(t) = 40 + 4t
(A + B)(t) = 40 + 2t
(A − B)(t) = −10 + 2t
(A − B)(t) = −10 + 4t

Answers

The expression that is useful to determine the total amount of credit cards opened at the two locations is (A + B)(t) = 40 + 2t so option (B) is correct.

What is an expression?

A mixture of variables, numbers, addition, subtraction, multiplication, and division are called expressions.

A statement expressing the equality of two mathematical expressions is known as an equation.

As per the given,

The amount in location A is given as

A(t) = 15 + 3t

The amount in location B is given as

B(t) = 25 − t

The total amount combined between A and B is given as,

(A + B)(t) = 15 + 3t + 25 - t

(A + B)(t) = 15 + 25 + 3t - t

(A + B)(t) = 40 + 2t

Hence "The expression that is useful to determine the total amount of credit cards opened at the two locations is (A + B)(t) = 40 + 2t".

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sean earns $300 in a regular work week. A regular work week for sean consists of 5 work days with 8 hours a day. How much money does sean earn each hour

Answers

Solution:

According to the problem, a regular work week consists of 5 work days with 8 hours a day. This is equivalent to say:

5 x 8 hours every regular work week.

That is:

40 hours every regular work week

then, the money earned per hour is:

[tex]\frac{300\text{ }dollars}{40\text{ hours}}\text{ = 7.5 dollars per hour}[/tex]

then we can conclude that the correct answer is:

$7.5

2. I want to find the area of a basketball free throw lane, and I am trying to use this formula:
Area =(6 ft) + (12 ft)(19 ft).
Clearly explain how you can tell this is wrong without knowing any area formulas.

Answers

Answer:

Down Here

Step-by-step explanation:

Hello!

Area is measured in units². Here, the units are feet, so the area would be written as feet² (ft²).

Simplifying the expression:

6ft + (12ft)(19ft)6ft + 228ft²

You can't simplify this because the units are not the same, and this can't measure area because all units of the area have to be squared. The term 6ft has non-squared units.

help meeeeeeeeee pleaseee !!!!!

Answers

The value of the composite function is determined as: (g o f)(5) = 6.

How to Determine the Value of a Composite Function?

To determine the value of a composite function, first evaluate the inner function by plugging in the value of x given, then use the output of the inner function as an input to evaluate the outer function.

Given the following:

f(x) = x² - 6x + 2

g(x) = -2x

Therefore:

(g o f)(5) = g(f(5))

Find f(5):

f(5) = (5)² - 6(5) + 2

f(5) = 25 - 30 + 2

f(5) = -3

Find g(f(5)) by substituting x = -3 into g(x) = -2x:

g(f(5)) = -2(-3)

g(f(5)) = 6

Therefore, (g o f)(5) = 6.

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Solve by using a proportion. Round answers to the nearest hundredth if necessary. 1. You jog 3.6 miles in 30 minutes. At that rate, how long will it take you to jog 4.8 miles? 2. You earn $33 in 8 hours. At that rate, how much would you earn in 5 hours?

Answers

EXPLANATION

Let's see the facts:

rate ---> 3.6 miles / 30 minutes

The unit rate is:

Unit rate = 0.12 miles/minute

Now, dividing the needed 4.8 miles by the unit rate will give us our desired number:

Time= 4.8 miles/ 0.12miles/minute = 40 minutes

The answer is 40 minutes.

Add the equation below:-9p=3p + 18Hint: We can isolate the variable by dividing each side by factors that don't contain the variable.

Answers

We have the next given equation:

[tex]9p=3p+18[/tex]

Now, we can subtract both sides by 3p:

[tex]\begin{gathered} 9p-3p=3p-3p+18 \\ 6p=18 \end{gathered}[/tex]

Then, divide both sides by 6:

[tex]\begin{gathered} \frac{6p}{6}=\frac{18}{6} \\ p=3 \end{gathered}[/tex]

Hence, the answer is p=3

hello I need help answering this homework question please thank you

Answers

Solution:

Case: Area

Given: A house to be painted

Method/ Final answers

a) Find the area of the garage to be painted.

(i) Front.

A = l X w - Garage door area

l= 15 ft, b= 10-4 gives 6ft

A= 15 X 6 - (10 X 7)

A= 90 - 70

A= 20 square feet

ii) Side

A= l X w

A= 6 X 5

A= 30 square feet

iii) The sum of areas

A= 20 + 30

A= 50 square feet

b) Area of the painted region around windows 5 and 6.

Since 12 in = 1 ft

Area of front door converted to feet is (20/3) ft by 3 ft

Areas of windows 5 and 6 converted to feet is 3 ft by (5/3) ft each

A= Area of space - (Area of front door + window 5 + window 6)

A= (30 X 10) - [(20/3) X 3 + 3 X (5/3) + 3 X (5/3)]

A= 300 - [20 + 5 + 5]

A= 300 - 30

A= 270 square feet.

c) Area of the painted region around windows 3

A= Total face - Area of window 3

A= (Rectangle + Parallelogram + Triangle) - Area of window 3

A= [(10 X 4) + (5 X 4) + (0.5 X 4 X 3)] - [1 X (5/3)]

A= [40+20+6] - [5/3]

A= 66 - (5/3)

A= 193/3 square feet

A= 63.33 square feet

d) Area of the region on the second floor with 2 rectangles and the region around window 4

i) region with rectangle 1 from left to right

A= 10 X (15- 6)

A= 10 X 9

A = 90

ii) region with rectangle 2 from left to right

A= 10 X (15- 6)

A= 10 X 9

A = 90

iii) Area of region around window 4

Area of space - area of window

A= 10 X (30-9-12) - [3 X (5/2)]

A= 10 X 9 - (15/2)

A= 82.5.

Total area= 90 + 90 + 82.5

= 262.5 square feet

e) Total area of the painted region (white)

262.5 + 63.33+ 270 + 50

= 645.83 square feet.

f) Additional question

The total cost if it cost $8 per sq ft

645.83 square feet X $8 per sq ft

=$5166.64

Ex5: The half-life of a certain radioactive isotope is 1430 years. If 24 grams are present now, howmuch will be present in 500 years?

Answers

[tex]A=A_0\cdot(\frac{1}{2})^{\frac{t}{h}}[/tex]

For the given situation:

[tex]\begin{gathered} A_0=24g \\ h=1430 \\ t=500 \\ \\ A=24(\frac{1}{2})^{\frac{500}{1430}} \\ \\ A=24(\frac{1}{2})^{\frac{50}{143}} \\ \\ A\approx18.83g \end{gathered}[/tex]Then, after 500 years there will be approxiomately 18.83 grams of the radioactive isotope

3. You have a bad cough and have to attend your little sister's choir concert. You take cough drops that contain 100 mg of menthol in each drop. Every minute, the amount of menthol in your body is cut in half. Write a funetion that models the amount of menthol in your body over time. Use x for minutes and y for the amount of menthol, in mg, remaining in your body It is safe to take a new cough drop after the level of menthol in your body is less than 5 mg, How long will it be before you can take another cough drop?

Answers

We have the next information

100 mg of menthol

every minute the amount of menthol in your body is cut in half

we have the next variables

x= minutes

y= amount of menthol in mg remaining in your body

so the equation that can we model is

[tex]y=100(0.5)^x[/tex]

then we have that It is safe to take a new cough drop after the level of menthol in your body is less than 5 mg

y= 5mg

[tex]5=100(0.5)^x[/tex]

in order to know the time we need to solve the equation above

[tex]\begin{gathered} (0.5)^x=\frac{5}{100} \\ (0.5)^x=0.05 \end{gathered}[/tex]

then we isolate the x

[tex]x=4.32[/tex]

after 5 minutes you can take another cough drop

Data Set A has a Choose... interquartile range than Data Set B. This means that the values in Data Set A tend to be Choose... the median.

Answers

The median of the given data set will be 35.

What do we mean by media?In statistics and probability theory, the median is the number that separates the upper and lower half of a population, a probability distribution, or a sample of data. For a data set, it might be referred to as "the middle" value.

So, The variability metrics for each class are listed below:

The further classifications: Class A; Class B;

Range: 30 Range: 30IQR: 12.5 IQR: 20.5MAD: 7.2 MAD: 9.2

Greater variability in the data set is suggested by class B's wider interquartile range and mean absolute deviations.

Set A's median will be:

median = (20 + 32+ 36+ 37 + 50) / 5median = 175 / 5median = 35

Therefore, the median of the given data set will be 35.

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Solve using elimination.–2x − 7y = 9x − 7y = –15

Answers

The question wants us to solve the following system of equations by elimination:

[tex]\begin{gathered} -2x-7=9 \\ x-7y=-15 \end{gathered}[/tex]

Solution

[tex]\begin{gathered} -2x-7y=9\text{ (Equation 1)} \\ x-7y=-15\text{ (Equation 2)} \\ \\ \text{Subtract both equations} \\ -2x-7y-(x-7y)=9-(-15) \\ -2x-7y-x+7y=9+15 \\ -2x-x-7y+7y=24 \\ -3x=24 \\ \text{Divide both sides by -3} \\ -\frac{3x}{-3}=\frac{24}{-3} \\ \\ \therefore x=-8 \\ \\ \text{Substitute the value for x into Equation 1}.\text{ This will help us find y.} \\ -2x-7y=9 \\ -2(-8)-7y=9 \\ 16-7y=9 \\ \text{Subtract 16 from both sides} \\ -7y=9-16 \\ -7y=-7 \\ \text{Divide both sides by -7} \\ -\frac{7y}{-7}=-\frac{7}{-7} \\ \\ \therefore y=1 \end{gathered}[/tex]

Answer

The answer to the system of equations is:

x = -8

y = 1

y varies inversely as x. y=12 when x=7. Find y when x=2

Answers

We write as an inverse proportion first then make an equation by multiplying by k:

[tex]y=\frac{k}{x}\Rightarrow k=x\times y[/tex]

Find the value of k:

[tex]k=7\times12=84[/tex]

Then, when x = 2, y is:

[tex]y=\frac{84}{2}=42[/tex]

Answer: y = 42

David is laying tiles on his kitchen floor. His kitchen measures 16 feet by 20 feet Each tile is a square that measures 2 feet by 2 feet (a) What is the area of his kitchen floor? (b) How many tiles will David need to purchase to cover the floor? One Tile 2 ft 2 ft

Answers

(a) To find the area of the kitchen floor, we just have to multiply

[tex]A=16ft\times20ft=320ft^2[/tex]

(b) To find the number of tiles needed, we have to find the area of each tile, which is

[tex]A_{\text{tile}}=2ft^{}\times2ft^{}=4ft^2[/tex]

Then, we divide the total area of the kitchen floor by the area of each tile.

[tex]n=\frac{320ft^2}{4ft^2}=80[/tex]

Hence, David will need 80 tiles to cover the floor.

A cone with radius 6 feet and height 15 feet is shown.6ftEnter the volume, in cubic feet, of the cone. Round youranswer to the nearest hundredth.

Answers

EXPLANATION:

Given;

We are given a cone with the following dimensions;

[tex]\begin{gathered} Dimensions: \\ Radius=6ft \\ Height=15ft \end{gathered}[/tex]

Required;

We are required to calculate the volume of the cone with the given dimensions.

Step-by-step solution;

To solve this problem, we would take note of the formula of the volume of a cone;

[tex]\begin{gathered} Volume\text{ }of\text{ }a\text{ }cone: \\ Vol=\frac{\pi r^2h}{3} \end{gathered}[/tex]

We can now substitute and we'll have;

[tex]Vol=\frac{3.14\times6^2\times15}{3}[/tex][tex]Vol=3.14\times36\times5[/tex][tex]Vol=565.2[/tex]

Therefore, the volume of the cone is,

ANSWER:

[tex]Volume=565.2ft^3[/tex]

inserted a picture of the question, can you just answer the question and not ask a lot of questions yes i’m following

Answers

Step-by-step explanation:

A nonagon has 9 sides, so a regular nonagon will have vertices that are 40° apart as measured from the center. It has 9-fold rotational symmetry,

so the figure will be identical to the original when rotated multiples of 360°/9 = 40°.

[tex]\frac{360}{9}=40[/tex]

Therefore the degrees will a nonagon have rotational symmetry

Hene the correct answer is Option B

Can you help me find the answer to this equation

Answers

Given:

K is the mid-point of FG and L is the mid-point FH.

To find:

The measure of Angle H

Step-by-step solution:

As we are given that K and L are the midpoints of FG and FH respectively.

According to the mid-point theoram,

The segment that connects the mid-point of two sides, is parallel to the third side of that triangle.

Thus,

KL||GH and ∠KLF = ∠GHL

As both of these angles are corresponding angles.

So we can say that m∠H = 85 degrees.

The following statements "If you are wearing a helmet, you are riding a bike." and "If you are not riding a bike, you are not wearing a helmet." are an example of a _____ statement.Select one:a.inverseb.conversec.contrapositive

Answers

Given:

The given statements are,

"If you are wearing a helmet, you are riding a bike."

"If you are not riding a bike, you are not wearing a helmet."

Required:

To identify the kind of statements.

Explanation:

We have the given statement:

"If you are wearing a helmet, you are riding a bike."

Here, p : you are wearing a helmet

q : you are riding a bike

Thus, taking negation of both the parts of the statement as follows:

If not q, then not p.

Hence, the statement formed is,

"If you are not riding a bike, you are not wearing a helmet."

This is the contrapositive statement.

Final Answer:

Given statements are an example of contrapositive.

Amanda y Pedro realizaron queques iguales, Lurdes se comió 2/4 partes, Amanda 2/3 y Pedro 3/4, quien comió menos?

Answers

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Meg owes the bank more than $15. Use , or = to make the statement true. Meg's account value ? -$15 2 What is the value of point A? ? How far is point A from 0 (absolute value)? || HAR

Answers

Answer; Meg's account is < - $15

Meg is owing the bank more than -$15

This implies that, the amount she is owing is more that -$15

The amount she is owing the bank could be -$16, -$17

Therefore, her current back account is less than -$15

Meg's account value is < -$15

At an all-you-can-eat barbeque fundraiser, adults pay $6 for a dinner and children pay $4 for a dinner. 212 people attend and you raise $1,128. What is the total number of adults and the total number of children attending? A)140 adults and 72 children B)72 adults and 140 children C)142 adults and 70 children D)70 adults and 142 children

Answers

A)140 adults and 72 children

Explanation

Step 1

Let x represents the number of childrend attending

Let y represents the number of adults attending

then

total cost for the children=4x

total cost for the adults=6y

if you raise 1128,

[tex]4x+6y=1128\text{ Equation(1)}[/tex]

Now, 212 people attend,Hence

[tex]x+y=\text{212 Equation(2)}[/tex]

Step 2

solve for x and y

a)isolate x in equation (2), then replace in equation (1)

[tex]\begin{gathered} x+y=212 \\ x=212-y \\ \text{now, replace} \\ 4x+6y=1128 \\ 4(212-y)+6y=1128 \\ 848-4y+6y=1128 \\ 2y+848=1128 \\ \text{subtract 848 in both sides} \\ 2y+848-848=1128-848 \\ 2y=280 \\ d\text{ivide boths ides by 2} \\ \frac{2y}{2}=\frac{280}{2} \\ y=140 \end{gathered}[/tex]

it means 140 adults are attending

b)replace y=140 in equatin (2) to find x

[tex]\begin{gathered} x+y=212 \\ x+140=212 \\ \text{subtract 140 in both sides} \\ x+140-140=212-140 \\ x=72 \end{gathered}[/tex]

so, the number of children is 72 and 72 children

1. Find all real solutions to each equation. (a) x(2x − 5) = 1

Answers

Use the distributive property to expand the parenthesis:

[tex]x(2x-5)=2x^2-5x[/tex]

Then:

[tex]undefined[/tex]

A model rocket is launched with an initial upward velocity of 156 ft/s. The rocket's height h (In feet) after t seconds is given by the following.
h=156t-16t²
Find all values of t for which the rocket's height is 60 feet.
Round your answer(s) to the nearest hundredth.
(If there is more than one answer, use the "or" button.)
Explanation
Check
ground
t = 0 seconds
☐or D
X
5
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I need help

Answers

The quadratic equation that gives the height of the rocket, h = 156·t - 16·t² is evaluated at h = 60 feet to give the two times the rocket's height is 60 feet as 0.40 seconds and 9.35 seconds.

What is a quadratic equation?

A quadratic equation is an equation of the second degree that can be expressed in the form; a·x² + b·x + c = 0, where the letters, a, and b represents the coefficients of x and c is a constant.

The initial velocity of the rocket = 156 ft./s upwards

The given equation of the rocket is: h = 156·t - 16·t²

The times when the rocket height is 60 feet are found by plugging in the value h = 60, in the equation of the vertical height of the rocket as follows:  

h = 60 = 156·t - 16·t²

156·t - 16·t² - 60 = 0

4·(39·t - 4·t² - 15) = 0

Therefore: [tex]39\cdot t - 4\cdot t^2 - 15 = \dfrac{0}{4} =0[/tex]

39·t - 4·t² - 15 = 0

-4·t² + 39·t - 15 = 0

From the quadratic formula which is used to solve the quadratic equation of the form; f(x) = a·x² + b·x + c, is presented as follows;

[tex]x = \dfrac{-b\pm\sqrt{b^2-4\cdot a \cdot c} }{2\cdot a}[/tex]

The solution of the equation, -4·t² + 39·t - 15 = 0, is therefore:

[tex]t = \dfrac{-39\pm\sqrt{(39)^2-4\times (-4) \times (-15)} }{2\times (-4)}= \dfrac{-39\pm\sqrt{1281} }{-8}[/tex]

Therefore, when the height of the rocket is 60 feet, the times are: [tex]t = \dfrac{-39-\sqrt{1281} }{-8}\approx 9.35[/tex] and [tex]t = \dfrac{-39+\sqrt{1281} }{-8}\approx 0.40[/tex]

The times when the height of the rocket is 60 feet, the times are:

t ≈ 9.35 s, and t ≈ 0.40 s

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14#An ecologist randomly samples 12 plants of a specific species and measures their heights. He finds that this sample has a mean of 14 cm and a standard deviation of 4 cm. If we assume that the height measurements are normally distributed, find a 95% confidence interval for the mean height of all plants of this species. Give the lower limit and upper limit of the 95% confidence interval. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.)Lower limit:Upper limit:

Answers

Answer:

Lower limit: 11.7 cm

Upper limit: 16.263

Explanation:

The formula to find the lower and upper limits of the confidence interval (given the data is normally distributed) is :

[tex]CI=\mu\pm Z^*\frac{\sigma}{\sqrt{n}}[/tex]

Where:

• μ = sample mean

,

• σ = sample standard deviation

,

• Z* = critical value of the z-distribution

,

• n = is the sample size

In this case:

• μ = 14cm

• σ = 4cm

,

• n = 12

The critical value of the z-distribution for a confidence interval of 95% is Z* = 1.96

Now, we can use the formula above to find the upper and lower limit:

[tex]CI=14\pm1.96\cdot\frac{4}{\sqrt{12}}=14\pm\frac{98\sqrt{3}}{75}=\frac{1050\pm98\sqrt{3}}{75}[/tex]

Thus:

[tex]Lower\text{ }limit=\frac{1050-98\sqrt{3}}{75}\approx11.736cm[/tex][tex]Upper\text{ }limit=\frac{1050-98\sqrt{3}}{75}\approx16.263cm[/tex]

Rounded to one decimal:

Lower limit: 11.7cm

Upper limit: 16.3cm

I need help with my math homework question please. Plus it has a second part of the question

Answers

The given quadratic equation is

y = - x^2 + 25

a) The leading coeffiecient is the coefficient of the term with the highest exponent. Thus, the leading coefficient is the coefficient of x^2.

Leading coefficient = - 1

Since the leading coefficient is negative, the graph would open downwards. Thus, the correct option is

Down

b) The standard form of a quadratic equation is

y = ax^2 + bx + c

By comparing both equations,

a = - 1

b = 0

c = 25

The formula for calculating the x coordinate of the vertex of the graph is

x = - b/2a

By substituting the given values,

x = - 0/2 * - 1 = 0

We would calculate the y coordinate of the vertex by substituting x = 0 into the original equation. We have

y = - 0^2 + 25

y = 25

The coordinate of the vertex is (0, 25)

c) To find the x intercepts, we would equate the original equation to zero and solve for x. We have

- x^2 + 25 = 0

x^2 = 25

Taking the square root of both sides,

x = square root of 25

x = ± 5

Thus, the x intercepts are

(5, - 5)

d) The y intercept is the value of y when x = 0

Substituting x = 0 into the orignal equation,

y = - 0^2 + 25

y = 25

y intercept = (0, 25)

e) We would find another point on the graph. Let us substitute x = 6 into the equation. We have

y = - (6)^2 + 25 = - 36 + 25

y = - 11

We would plot (6, - 11) and (0, 25) on the graph. The graph is shown below

Writing an Equation Assume that the ball rebounds the same percentage on each bounce. Using the initial drop height and the height after the first bounce, find the common ratio,r.Note: Round r to three decimal places. Use this formula:common ratio = height on first bounce/initial heightheight on first bounce = 54 in Dropped from 72in (6 feet)

Answers

The common ratio = 0.750 (3 decimal places)

Explanation:

Initial drop height = 72 inches

Height after the first bounce = 54 inches

common ratio = r = height on first bounce/initial height

r = 54/72

r = 0.75

The common ratio = 0.750 (3 decimal places)

I just need to know if You just have to tell me if the circles are open or closed.

Answers

Solution

- The solution is given below:

[tex]\begin{gathered} y-2<-5 \\ y-2>5 \\ \\ \text{ Add 2 to both sides} \\ \\ y<-5+2 \\ y<-3 \\ \\ y-2>5 \\ y>5+2 \\ y>7 \end{gathered}[/tex]

- Thus, we have:

[tex]\begin{gathered} y<-3 \\ or \\ y>7 \end{gathered}[/tex]

- Thus, the plot is:

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