5. The number of hours spent in an airplane on a single flight is recordedon a dot plot. The mean is 5 hours. The median is 4 hours. The IQR is 3hours. The value 26 hours is an outlier that should not have been includedin the data. When 26 is removed from the data set, calculate the following(some values may not be used):*H0 2 4 6 8 10 12 14 16 18 20 22 24 26 28number of hours spent in an airplane1.4 hours1.5 hours3 hours3.5 hoursWhat is themean?OWhat is themedian?оOOWhat is the IQR?OOOO
Answers
Solution
Since the outlier that is 26 has been removed
We will work with the remaining
Where X denotes the number of hours, and f represent the frequency corresponding to eaxh hours
We find the mean
The mean (X bar) is given by
[tex]\begin{gathered} mean=\frac{\Sigma fx}{\Sigma f} \\ mean=\frac{1(2)+2(2)+3(3)+4(3)+5(2)+6(2)}{2+2+3+3+2+2} \\ mean=\frac{2+4+9+12+10+12}{2+2+3+3+2+2} \\ mean=\frac{49}{14} \\ mean=\frac{7}{2} \\ mean=3.5 \end{gathered}[/tex]We now find the median
Median is the middle number
Since the total frequency is 14
The median will be on the 7th and 8th term in ascending order
[tex]\begin{gathered} median=\frac{7th+8th}{2} \\ median=\frac{3+4}{2} \\ median=\frac{7}{2} \\ median=3.5 \end{gathered}[/tex]Lastly, we will find the interquartile range
The formula is given by
[tex]IQR=Q_3-Q_1[/tex]Where
[tex]\begin{gathered} Q_3=\frac{3}{4}(n+1)th\text{ term} \\ Q_1=\frac{1}{4}(n+1)th\text{ term} \end{gathered}[/tex]We calculate for Q1 and Q3
[tex]\begin{gathered} Q_1=\frac{1}{4}(n+1)th\text{ term} \\ \text{n is the total frequency} \\ n=14 \\ Q_1=\frac{1}{4}(14+1)th\text{ term} \\ Q_1=\frac{1}{4}(15)th\text{ term} \\ Q_1=3.75th\text{ term} \\ Q_1\text{ falls betwe}en\text{ the frequency 3 and 4 in ascending order} \\ \text{From the table above} \\ Q_1=2 \end{gathered}[/tex][tex]\begin{gathered} Q_3=\frac{3}{4}(n+1)th\text{ term} \\ Q_3=\frac{3}{4}(14+1)th\text{ term} \\ Q_3=\frac{3}{4}(15)th\text{ term} \\ Q_3=11.25th\text{ term} \\ \text{From the table above} \\ Q_3=5 \end{gathered}[/tex]Therefore, the IQR is
[tex]\begin{gathered} IQR=Q_3-Q_1 \\ IQR=5-2 \\ IQR=3 \end{gathered}[/tex]
Related Questions
The American Water Works Association reports that the per capita water use in a single-family home is 69 gallons per day. Legacy Ranch is a relatively new housing development. The builders installed more efficient water fixtures, such as low-flush toilets, and subsequently conducted a survey of the residences. Thirty-six owners responded, and the sample mean water use per day was 64 gallons with a standard deviation of 8.8 gallons per day.
At the .10 level of significance, is that enough evidence to conclude that residents of Legacy Ranch use less water on average?
What is the decision rule? (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)
Reject H0: µ ≥ 69 when the test statistic is less than ____.
Answers
a. The decision rule for this question would be to reject the null hypothesis if test statistic is less than the critical value.
b. The test statistic is given as: -3.4091
What is the hypothesis?We have the null hypothesis as
h0 : μ ≥ 69
The alternate hypothesis is
H1 : μ < 69
a. The decision rule would be to reject the null if the test statistic is greater than the critical value
at α = 0.10 the degree of freedom = 36 - 1 = 35
the critical value is -1.306
The test statistic calculation
[tex]t =\frac{ x - u}{s/\sqrt{n} }[/tex]
[tex]t = \frac{64-69}{8.8/\sqrt{36} }[/tex]
t = -3.4091
The decision rule would be to Reject H0: µ ≥ 69 when the test statistic is less than -1.306.
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What transaction occurs when an investor decides to liquidate assets?
A. buy
B. hold
C. sell
D. speculate
Answers
Answer:
What transaction occurs when an investor decides to liquidate assets?
A. buy
B. hold
(C. sell)
D. speculate
Step-by-step explanation:
I got a 5/5 on the test and i got the answer from a quizlet (:
Sell is the answer
The correct option is (C).
Given,
In the question:
What transaction occurs when an investor decides to liquidate assets?
Now, According to the question:
when an investor decides to liquidate assets.
when an investor decides to liquidate assets means he or she want to sell the property in the open market, in other words liquidate assets means
converting non- liquid assets into liquid assets.
In investing, liquidation occurs when an investor closes their position in an asset. Liquidating an asset is usually carried out when an investor or portfolio manager needs cash to re-allocate funds or rebalance a portfolio. An asset that is not performing well may also be partially or fully liquidated.
According to the statement
Therefore, Sell is the answer
The correct option is (C).
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a store donated 2 and 1/4 cases of cranes to a daycare center each case holds 24 boxes of crayons each box holds 8 crayons how many crayons did the center receive
Answers
Answer:
The center recieved 432 crayons
Explanation:
Given the following information:
There are 2 and 1/4 cases
Each case holds 24 boxes of crayons
Each box holds 8 crayons.
The number of crayons the center receive is:
8 * 24 * (2 + 1/4)
= 8 * 24 * (8/4 + 1/4)
= 192 * (9/4)
= 1728/4
= 432
Subtract and simplify the answer. 8/9 - 1/3
Answers
Solution
We want to simplify
[tex]\frac{8}{9}-\frac{1}{3}[/tex]Now
[tex]\begin{gathered} \frac{8}{9}-\frac{1}{3}=\frac{8}{9}-\frac{1\times3}{3\times3} \\ \frac{8}{9}-\frac{1}{3}=\frac{8}{9}-\frac{3}{9} \\ \frac{8}{9}-\frac{1}{3}=\frac{8-3}{9} \\ \frac{8}{9}-\frac{1}{3}=\frac{5}{9} \end{gathered}[/tex]Therefore, the answer is
[tex]\frac{5}{9}[/tex](c) Given that q= 8d^2, find the other two real roots.
Answers
Polynomials
Given the equation:
[tex]x^5-3x^4+mx^3+nx^2+px+q=0[/tex]Where all the coefficients are real numbers, and it has 3 real roots of the form:
[tex]x_1=\log _2a,x_2=\log _2b,x_3=\log _2c[/tex]It has two imaginary roots of the form: di and -di. Recall both roots must be conjugated.
a) Knowing the sum of the roots must be equal to the inverse negative of the coefficient of the fourth-degree term:
[tex]\begin{gathered} \log _2a+\log _2b+\log _2c+di-di=3 \\ \text{Simplifying:} \\ \log _2a+\log _2b+\log _2c=3 \\ \text{Apply log property:} \\ \log _2(abc)=3 \\ abc=2^3 \\ abc=8 \end{gathered}[/tex]b) It's additionally given the values of a, b, and c are consecutive terms of a geometric sequence. Assume that sequence has first term a1 and common ratio r, thus:
[tex]a=a_1,b=a_1\cdot r,c=a_1\cdot r^2[/tex]Using the relationship found in a):
[tex]\begin{gathered} a_1\cdot a_1\cdot r\cdot a_1\cdot r^2=8 \\ \text{Simplifying:} \\ (a_1\cdot r)^3=8 \\ a_1\cdot r=2 \end{gathered}[/tex]As said above, the real roots are:
[tex]x_1=\log _2a,x_2=\log _2b,x_3=\log _2c[/tex]Since b = a1*r, then b = 2, thus:
[tex]x_2=\log _22=1[/tex]One of the real roots has been found to be 1. We still don't know the others.
c) We know the product of the roots of a polynomial equals the inverse negative of the independent term, thus:
[tex]\log _2a_1\cdot2\cdot\log _2(a_1\cdot r^2)\cdot(di)\cdot(-di)=-q[/tex]Since q = 8 d^2:
[tex]\begin{gathered} \log _2a_1\cdot2\cdot\log _2(a_1\cdot r^2)\cdot(di)\cdot(-di)=-8d^2 \\ \text{Operate:} \\ 2\log _2a_1\cdot\log _2(a_1\cdot r^2)\cdot(-d^2i^2)=-8d^2 \\ \log _2a_1\cdot\log _2(a_1\cdot r^2)=-8 \end{gathered}[/tex]From the relationships obtained in a) and b):
[tex]a_1=\frac{2}{r}[/tex]Substituting:
[tex]\begin{gathered} \log _2(\frac{2}{r})\cdot\log _2(2r)=-8 \\ By\text{ property of logs:} \\ (\log _22-\log _2r)\cdot(\log _22+\log _2r)=-8 \end{gathered}[/tex]Simplifying:
[tex]\begin{gathered} (1-\log _2r)\cdot(1+\log _2r)=-8 \\ (1-\log ^2_2r)=-8 \\ \text{Solving:} \\ \log ^2_2r=9 \end{gathered}[/tex]We'll take the positive root only:
[tex]\begin{gathered} \log _2r=3 \\ r=8 \end{gathered}[/tex]Thus:
[tex]a_1=\frac{2}{8}=\frac{1}{4}[/tex]The other roots are:
[tex]\begin{gathered} x_1=\log _2\frac{1}{4}=-2 \\ x_3=\log _216=4 \end{gathered}[/tex]Real roots: -2, 1, 4
I need help solving an optimization math problem please :)
Answers
Answer:
Explanation:
Let the side opposite the river = x
Let the adjacent side to the river = y
Solve for x:
A
+79
X
Answers
Answer: -11
Step-by-step explanation: 66+46=112
180-112=68
79+?=68
79+-11=68
For f(x)=x^2 and g(x)=x^2+9, find the following composite functions and state the domain of each.
(a) f.g (b) g.f (c) f.f (d) g.g
Answers
The composite functions in this problem are given as follows:
a) (f ∘ g)(x) = x^4 + 18x² + 81.
b) (g ∘ f)(x) = x^4 + 9.
c) (f ∘ f)(x) = x^4.
d) (g ∘ g)(x) = x^4 + 18x² + 90.
All these functions have a domain of all real values.
Composite functionsFor composite functions, the outer function is applied as the input to the inner function.
In the context of this problem, the functions are given as follows:
f(x) = x².g(x) = x² + 9.For item a, the composite function is given as follows:
(f ∘ g)(x) = f(x² + 9) = (x² + 9)² = x^4 + 18x² + 81.
For item b, the composite function is given as follows:
(g ∘ f)(x) = g(x²) = (x²)² + 9 = x^4 + 9.
For item c, the composite function is given as follows:
(f ∘ f)(x) = f(x²) = (x²)² = x^4.
For item d, the composite function is given as follows:
(g ∘ g)(x) = g(x² + 9) = (x² + 9)² + 9 = x^4 + 18x² + 90.
None of these functions have any restriction on the domain such as fractions or even roots, hence all of them have all real values as the domain.
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hello and thank you for helping me and this is a trigonometry question bit for the question has give exact value and it won't accept decimals as an answer and thank you for your time.
Answers
1) In this question let's calculate the sin(θ) and cos(θ)
Given that
[tex]\begin{gathered} \text{If }\sin (\theta)=\frac{5\pi}{4} \\ \sin (\theta)\text{ }\Rightarrow\sin (\frac{5\pi}{4})\text{ }=-\frac{\sqrt[]{2}}{2} \\ \cos (\frac{5\pi}{4})=-\frac{\sqrt[]{2}}{2} \end{gathered}[/tex]2) In this question, we're calculating the value of the sine and the cosine in radians.
We must remember that 5π/4 ⇒ to 225º, and that it's in the Quadrant III
If we subtract
225 -180 =45 So the sine of 5π/4 is -√2/2 and the cosine (5π/4 ) = -√2/2
2.3) The sign of the Quadrant
Since 225º is in Quadrant III both results are negative ones.
probleme 1-2 show two Parallel lines and a transversal. Find the values of x
Answers
From the blurry picture shown, we can concur that:
x and 123.25 degree angle are interior corresponding angles.
They add up to 180 degrees, thus we can write the equation:
[tex]123.25\degree+x\degree=180\degree[/tex]We can now easily solve for x:
[tex]\begin{gathered} 123.25\degree+x\degree=180\degree \\ x\degree=180-123.25 \\ x=56.75\degree \end{gathered}[/tex]The solution:
[tex]x=56.75\degree[/tex]Solve for a side in right triangles. AC = ?. Round to the nearest hundredth
Answers
The length of segment AC is 2.96 units
How to determine the side length AC?From the question, the given parameters are
Line segment AB = 7 units
Angle A = 65 degrees
The line segment AC can be calculated using the following cosine ratio
cos(Angle) = Adjacent/Hypotenuse
Where
Adjacent = Side length AC
Hypotenuse = Side length AB
So, we have
cos(65) = AC/AB
This gives
cos(65) = AC/7
Make AC the subject
AC =7 * cos(65)
Evaluate
AC = 2.96
Hence, the side length AC has a value of 2.96 units
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60 cars to 24 cars The percent of change is
Answers
We can calculate the percent of change by means of the following formula:
[tex]change=\frac{x2-x1}{x1}\times100[/tex]Where x2 is the new value and x1 is the original value.
In this case, we go from 60 to 24, then the original value (x1) was 60 and the new value (x2) is 24, by replacing these values into the above equation, we get:
[tex]change=\frac{24-60}{60}\times100=-60[/tex]Then, the percent of change equals -60%
system of equationsb+c= -55b-c= 17
Answers
Let's solve the system of equations:
b + c = - 55
b - c = 17
Step 1: Let's isolate b on the first equation:
b + c = - 55
b = - 55 - c
Step 2: Let's solve for c on the second equation, substituting b:
b - c = 17
-55 - c - c = 17
-55 - 2c = 17
Adding 55 at both sides:
-2c - 55 + 55 = 17 + 55
-2c = 72
Dividing by - 2 at both sides:
-2c/-2 = 72/-2
c = -36
Step 3: Let's solve for b on the first equation, susbtituting c:
b + c = - 55
b + (-36) = - 55
b - 36 = - 55
Adding 36 at both sides:
b - 36 + 36 = - 55 + 36
I think you are ready to finish and calculate the value for b.
In scalene triangle ABC shown in the diagram below, m2C = 90°.B.Which equation is always true?sn A = sin Bcos sn A = cos BCanAB4 5 678 9 1011
Answers
inNote: To know which equation is true, then we will have to TEST for each of the choices we are to pick from.
From the tirangle in the image.
[tex]\begin{gathered} 1)\sin \text{ A =}\frac{\text{ Opp}}{\text{Hyp}}\text{ = }\frac{a}{c} \\ \cos \text{ B = }\frac{\text{ADJ}}{\text{HYP}}\text{ = }\frac{a}{c} \\ So\text{ from the above, we can s}ee\text{ that: SinA = Cos B :This mean the choice are equal} \\ \end{gathered}[/tex][tex]\begin{gathered} 2)\text{ To test for the second choice we have..} \\ \text{ Cos A = Cos B} \\ \text{for Cos A =}\frac{\text{Adj}}{\text{Hyp}}\text{ =}\frac{b}{c} \\ \\ \text{for Cos B = }\frac{Adj}{\text{Hyp}}\text{ = }\frac{a}{c} \\ \text{from here we can s}ee\text{ that Cos A }\ne\text{ Cos B : meaning Cos A is not equal to Cos B} \\ \end{gathered}[/tex]3) To test for the third choice: Sin A = Cos A
[tex]\begin{gathered} \sin \text{ A=}\frac{opp}{\text{Hyp}}\text{ = }\frac{a}{c} \\ \cos \text{ A = }\frac{Adj}{\text{Hyp}}\text{ = }\frac{b}{c} \\ we\text{ can s}ee\text{ that sinA }\ne\text{ cos }A,\text{ This mean they are not equal} \end{gathered}[/tex][tex]\begin{gathered} 4)\text{ To test if: tan A = sin B} \\ \text{ }tan\text{ A = }\frac{opp}{\text{Adj}}\text{ = }\frac{a}{b} \\ \\ \text{ sin B = }\frac{Opp}{\text{Hyp}}\text{ = }\frac{b}{c} \\ so\text{ from what we have, w can s}ee\text{ that tan A }\ne\text{ sinB: Meaning they are not equal.} \end{gathered}[/tex]Meaning the first choice is the answer that is sin A = CosB
Help me with my schoolwork what is the slope of line /
Answers
The two points given on the line are
[tex]\begin{gathered} (x_1,y_1)\Rightarrow(-2,9) \\ (x_2,y_2)\Rightarrow(6,1) \end{gathered}[/tex]The slope of line that passes through (x1,y1) and (x2,y2) is gotten using the formula below
[tex]\begin{gathered} m=\frac{\text{change in y}}{\text{change in x}} \\ m=\frac{y_2-y_1}{x_2-x_1} \end{gathered}[/tex]By substituting the values, we will have
[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{1-9}{6-(-2)} \\ m=-\frac{8}{6+2} \\ m=-\frac{8}{8} \\ m=-1 \end{gathered}[/tex]Therefore,
The slope of the line = -1
in a sale normal prices are reduced by 15%. The sale price of a CD player is £102. work out the normal price of the CD player
Answers
The normal price for the CD player is $117.30
How to calculate the value?Since the normal prices are reduced by 15%, the percentage for the normal price will be:
= 100% + 15%
= 115%
Also, the sale price of a CD player is £102.
Therefore, the normal price will be:
= Percentage for normal price × Price
= 115% × $102
= 1.15 × $102
= $117.30
The price is $117.30.
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Martin earns $7.50 per hour proofreading ads per hour proofreading ads at a local newspaper. His weekly wage can. e found by multiplying his salary times the number of hours h he works.1. Write an equation.2. Find f(15)3. Find f (25)
Answers
If Martin earns 7.50 per hour (that is h), then the equation for his weekly wage can be expressed as;
[tex]\begin{gathered} (A)f(h)=7.5h \\ (B)f(15)=7.5(15) \\ f(15)=112.5 \\ (C)f(25)=7.5(25) \\ f(25)=187.5 \end{gathered}[/tex]Therefore, answer number A shows the equation for his salary
Answer number 2 shows his salary at 15 hours ($112.5)
Answer number 3 shows his salary at 25 hours ($187.5)
. Estimate the area of a parallelogram with a base of 3 ¼ yards and a height of 5 ½ yards.
Answers
We are given the dimensions of a parallelogram and are asked to estimate its area
Recall that the area of a parallelogram of base b and height h is given by the formula
[tex]A=b\cdot h[/tex]So the area of the parallelogram would be
[tex]3\frac{1}{4}\cdot5\frac{1}{2}[/tex]as 3 1/4 and 5 1/2 are mixed numbers, we need to transform them to fractions
Recall that given a mixed number of the form
[tex]a\frac{b}{c}[/tex]we can transform it into a fraction by multiplying the whole number by the denominator and adding the result to the numerator while leaving the denominator fixed. In this case, that is
[tex]a\frac{b}{c}=\frac{a\cdot c+b}{c}[/tex]So, applying this formula to both numbers, we get
[tex]3\frac{1}{4}=\frac{3\cdot4+1}{4}=\frac{13}{4}[/tex]and
[tex]5\frac{1}{2}=\frac{5\cdot2+1}{2}=\frac{11}{2}[/tex]so the area of the parallelogram would be
[tex]\frac{13}{4}\cdot\frac{11}{2}=\frac{143}{8}\approx18[/tex]so the area of the parallelogram is approximately 18 square yards
In the xy-plane, line n passes through point (0,0) and has a slope of 4. If line n also passes through point (3,a), what is the value of a?
Answers
Select all the situations in which a proportional relationship is described.
Jackson saves $10 in the first month and $30 in the next 3 months.
Mia saves $8 in the first 2 months and $4 in the next month.
Piyoli spends $2 in the first 2 days of the week and $5 in the next 5 days.
Robert spends $2 in the first 3 days of the week and $5 in the next 4 days.
Answers
Answer:
Jackson saves $10 in the first month and $30 in the next 3 months.
Mia saves $8 in the first 2 months and $4 in the next month.
Piyoli spends $2 in the first 2 days of the week and $5 in the next 5 days.
Step-by-step explanation:
A proportional relationship is one that has a constant of proportionality.
In this case, the correct options are Mia, Piyoli, and Robert.
Give the equation of the line parallel to a line through (-3, 4) and (-5, -6) that passes through the origin. y = 5x y = 5x + 1 y=-1/5x + 1 y = -1/5x y
Answers
To solve for the equation of the line parallel :
[tex]\begin{gathered} (-3,4)\Longrightarrow(x_1,y_1) \\ (-5,-6)\Longrightarrow(x_2,\text{y}_2) \end{gathered}[/tex]For parallel line equation:
Slope-intercept form: y=mx+b, where m is the slope and b is the y-intercept
First let's find the slope of the line.
To find the slope using two points, divide the difference of the y-coordinates by the difference of the x-coordinates.
[tex]\begin{gathered} \text{slope =}\frac{y_2-y_1}{x_2-x_1} \\ \text{slope}=\frac{-6-4}{-5--3} \\ \text{slope=}\frac{-10}{-5+3}=\frac{-10}{-2} \\ \text{slope =5} \end{gathered}[/tex]Slope= 5
[tex]\begin{gathered} y=mx+c \\ y=5x+c \\ \text{where c = y-intercept} \end{gathered}[/tex]The y-intercept is (0, b). The equation passes through the origin, so the y-intercept is 0.
[tex]\begin{gathered} y=5x+0 \\ y=5x \end{gathered}[/tex]Hence the
A chocolate factory has a goal to produce10121012pounds of chocolate frogs per day. If the machines operate for712712hours per day making215215pounds of chocolate frogs per hour, will the chocolate factory make it’s goal?The chocolate factory meet their goal with the total being10121012pounds of chocolate frogs produced.
Answers
First, rewrite all the mixed fractions as impropper fractions:
[tex]\begin{gathered} 10\frac{1}{2}=10\times\frac{2}{2}+\frac{1}{2}=\frac{20}{2}+\frac{1}{2}=\frac{21}{2} \\ \\ 7\frac{1}{2}=7\times\frac{2}{2}+\frac{1}{2}=\frac{14}{2}+\frac{1}{2}=\frac{15}{2} \\ \\ 2\frac{1}{5}=2\times\frac{5}{5}+\frac{1}{5}=\frac{10}{5}+\frac{1}{5}=\frac{11}{5} \end{gathered}[/tex]Next, multiply the rate of chocolate production over time by the the operating time of the machines to find the total amount of pounds of chocolate frogs produced in one day:
[tex]7\frac{1}{2}\times2\frac{1}{5}=\frac{15}{2}\times\frac{11}{5}=\frac{15\times11}{2\times5}=\frac{3\times11}{2}=\frac{33}{2}=16\frac{1}{2}[/tex]Then, the chocolate factory can produce 16 1/2 pounds of chocolate frogs per day.
Since 16 1/2 is greater than 10 1/2, then the chocolate factory will meet their goal with the total being over 10 1/2 pounds of chocolate frogs produced.
Seventh grade > X.9 Reflections over the x- and y-axes: find the coordinates TF8 You have prizes to reveal The point D(-5, -3) is reflected over the y-axis. What are the coordinates of the resulting point, D'?
Answers
Answer
The coordinates of the resulting point, D' = (5, -3)
Explanation
When a given point with coordinates A (x, y) is reflected over the y-axis, the y-coordinate remains the same and the x-coordinate takes up a negative in front of it. That is, A (x, y) changes after being reflected across the y-axis in this way
A (x, y) = A' (-x, y)
So, for this question where the coordinate is D (-5, -3). it changes in the manner,
D (-5, -3) = D' (-(-5), -3) = D' (5, -3)
Hope this Helps!!!
StatusExam9 ft.15 ft.The volume ofthe figure iscubic feet.15 ft.15 ft.
Answers
Step 1:
The figure is a composite figure with a square base pyramid and a cube.
Step 1:
The volume of the composite shape is the sum of the volume of a square base pyramid and a cube.
[tex]\text{Volume = L}^3\text{ + }\frac{1}{2}\text{ base area }\times\text{ height}[/tex]Step 3:
Given data
Cube
Length of its sides L = 15 ft
Square base pyramid
Height h = 9 ft
Length of the square base = 15 ft
Step 4:
Substitute in the formula.
[tex]\begin{gathered} \text{Volume = 15}^3\text{ + }\frac{1}{3}\text{ }\times15^2\text{ }\times\text{ 9} \\ \text{= 3375 + 675} \\ =4050ft^3 \end{gathered}[/tex]Find the x-coordinate for the point of intersection by using the equations method of solving. Show all the work. f (x)=2x+6g(x)= -3x+1
Answers
y = 2x + 6
y = -3x + 1
Equality
2x + 6 = -3x + 1 6 is in the left side and is positive so we substract 6 in both sides
2x +6 - 6 = -3x + 1 - 6
Simplify
2x = -3x - 5
Add -3x in both sides
2x + 3x = -3x + 3x - 5
Simplify
5x = -5
2x + 3x = 1 - 6
5x = -5
x = -5/5
This is the x-coordinate
x = -1
Solve the following inequality. Graph the solution set and then write it in interval notation .
Answers
Given:
-2x ≥ 6
Solve for x
Divide both sides by -2
-2x/-2 ≤ 6/-2
x ≤ -3
Graph:
Interval notation (-∞, -3 ]
Determine which of the lines are parallel and which of the lines are perpendicular. Select all of the statements that are true.
Line a passes through (-1, -17) and (3, 11).
Line b passes through (0,4) and (7,-5).
Line c passes through (7, 1) and (0, 2).
Line d passes through (-1,-6) and (1, 8).
Answers
Answers:
Line A is parallel to line D.
Line A is perpendicular to line C.
Line C is perpendicular to line D.
=====================================================
Explanation:
Let's use the slope formula to calculate the slope of the line through (-1,-17) and (3,11)
[tex](x_1,y_1) = (-1,-17) \text{ and } (x_2,y_2) = (3,11)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{11 - (-17)}{3 - (-1)}\\\\m = \frac{11 + 17}{3 + 1}\\\\m = \frac{28}{4}\\\\m = 7\\\\[/tex]
The slope of line A is 7
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Now let's find the slope of line B.
[tex](x_1,y_1) = (0,4) \text{ and } (x_2,y_2) = (7,-5)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{-5 - 4}{7 - 0}\\\\m = -\frac{9}{7}\\\\[/tex]
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Now onto line C.
[tex](x_1,y_1) = (7,1) \text{ and } (x_2,y_2) = (0,2)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{2 - 1}{0 - 7}\\\\m = \frac{1}{-7}\\\\m = -\frac{1}{7}\\\\[/tex]
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Lastly we have line D.
[tex](x_1,y_1) = (-1,-6) \text{ and } (x_2,y_2) = (1,8)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{8 - (-6)}{1 - (-1)}\\\\m = \frac{8 + 6}{1 + 1}\\\\m = \frac{14}{2}\\\\m = 7\\\\[/tex]
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Here's a summary of the slopes we found
[tex]\begin{array}{|c|c|} \cline{1-2}\text{Line} & \text{Slope}\\\cline{1-2}\text{A} & 7\\\cline{1-2}\text{B} & -9/7\\\cline{1-2}\text{C} & -1/7\\\cline{1-2}\text{D} & 7\\\cline{1-2}\end{array}[/tex]
Recall that parallel lines have equal slopes, but different y intercepts. This fact makes Line A parallel to line D.
Lines A and C are perpendicular to one another, because the slopes 7 and -1/7 multiply to -1. In other words, -1/7 is the negative reciprocal of 7, and vice versa. These two lines form a 90 degree angle.
Lines C and D are perpendicular for the same reasoning as the previous paragraph.
Line B unfortunately is neither parallel nor perpendicular to any of the other lines mentioned.
You can use a graphing tool like Desmos or GeoGebra to verify these answers.
In the diagram, MN is parallel to KL. What is the length of MN? K M 24 cm 6 cm 2 12 cm L O A. 6 cm O B. 18 cm O c. 12 cm D. 8 cm
Answers
To solve this question, we shall be using the principle of similar triangles
Firstly, we identify the triamgles
These are JKL and JMN
JKL being the bigger and JMN being the smaller
Mathematically, when two triangles are similar, the ratio of two of their corresponding sides are equal
Thus, we have it that;
[tex]\begin{gathered} \frac{JN}{MN}\text{ = }\frac{JL}{KL} \\ \\ \frac{6}{MN}=\text{ }\frac{18}{24} \\ \\ MN\text{ = }\frac{24\times6}{18} \\ MN\text{ = 8 cm} \end{gathered}[/tex]Inside. Make sure you don’t enter any spaces in your answers. This answer needs to be rounded to the nearest hundredth.
Answers
ANSWER
c = 14.14
EXPLANATION
To find the length of side AB, which is the hypotenuse of the right triangle ABC, we have to apply the Pythagorean Theorem,
[tex]AB^2=AC^2+BC^2[/tex]Replace the known values and solve for c,
[tex]c^2=10^2+10^2=100+100=200\Rightarrow c=\sqrt{200}\approx14.14[/tex]Hence, the value of c is 14.14, rounded to the nearest hundredth.
If mABC =(3x+3) and mDEF=(5x-33).Find the value of x
Answers
Let's begin by listing out the information given to us:
m∠ABC = 3x + 3
m∠DEF = 5x - 33
From the question, m∠ABC & m∠DEF are identical (have same properties)
m∠ABC = m∠DEF
3x + 3 = 5x - 33
Put like terms together (add 33 - 3x to both sides)
3x - 3x + 3 + 33 = 5x - 3x - 33 + 33
36 = 2x; 2x = 36
x = 18
