Which pairs of rational numbers are equivalent? Select all that apply.
A. 0.5 and -1 over 2
B. 0.7 and 7 over 10
C. 11 over 12 and 0.55
D. -1 over 50 and -0.02
E. 0.15 and 0.15​

Answer:

2. ) {25}^{2x} - 1 = {125}^{3x} + 4

=> {25}^{2x} = {125}^{3x} + 4 +1

=> {(5)^(2)}^{2x} = {(5)^(3)}^{3x} + {5}^{1}

=> {5}^{4x} = {5}^{9x} + {5}^{1}

=> 4x = 9x + 1

=> 4x - 9x = 1

=> - 5x = 1

=> x = 1/-5

Answer:

[tex]\frac{19}{6} =3\frac{1}{6}[/tex]

Step-by-step explanation:

Answer:

The answer is 3 1/6

Step-by-step explanation:

Answer:

True

Step-by-step explanation:

A Kilogram (kilo) is a unit of measuring metric mass. To perform this conversion you need to note that

1 Kilogram (kg) = 1000 grams

Which means that converting Kilogram to Grams would involve multiplying kilogram by 1000.

For example,  100 kg to grams = 100 * 1000 = 100,000 Grams

Answer:

The answer is C

Step-by-step explanation:

Answer:

I think that the answer is C I hope that this helpful

Step-by-step explanation:

C

Answer:

C. Car

Step-by-step explanation

Answer:

x = 4 and the expression when evaluated at x = 4, is 16.

Step-by-step explanation:

For this problem, we will simply look at the wording of the question and the given expression.  The question ask, "What would x be equal to if x equaled 4 in this: 4x?"

In the problem, we are already given that x = 4 in the following expression, so the answer to the question is x = 4.  If we were to evaluate the expression when x = 4, we would say the following:

If x = 4, then:

4x

= 4(4)

= 16

Hence, the expression would be 16 if evaluated when x = 4.

Cheers.

Answer:

The main flaw of this  proof is that it did not take into consideration the fact a 4 cent postage must be made with a 4 cent stamp  and a 3 cent postage  must be made with a 3 cent stamp but only looked at allocating the stamps based on the total amount of postage

Step-by-step explanation:

From the question we are told that

   The  proof is

Every postage stamp of three cents or more can be formed using just 3 cent and 4 cent stamps.  

 Basic Step : We can form postage of 3 cents with a single 3 cent stamp and we can form postage of 4 cents using a single four cents tamp

Inductive Step: Assume that we can form postage of j cents for all non negative integers j with j <= k using just 3 cent and 4-cent stamps. We can then form postage of k+1 cents by replacing one 3 cent stamp with a 4-cent or two 4-cent stamps with three 3-cent stamps

  FLAW IDENTIFICATION

Looking the inductive step, the statement ''by replacing one 3 cent stamp with a 4-cent or two 4-cent stamps with three 3-cent stamps '' means that it is possible to use a 3-cent stamp for postage that requires a 4 -cent stamp  and this is not correct according the statement in the basic step

Generally the main flaw of this  proof is that it did not take into  consideration the fact a 4 cent postage must be made with a 4 cent stamp  and a 3 cent postage  must be made with a 3 cent stamp but only looked at allocating the stamps based on the total amount of postage

Answer:

it is 237

Step-by-step explanation:

Answer: 237

Step-by-step explanation:

You do not to do order of operations for this problem

2 x 100 = 200

200 + 18 = 218

218 - 1 = 217

217 + 20 = 237

Hope it helps!

Answer:

(a) Tens=2

(b) Ten thousandths=8

(c) Tenths=4

(d) Ones=1

Step-by-step explanation:

The given number is 921.430865.

The place value names before the decimal point in the number are

[tex]\cdots-\text{Thousands}-\text{Hundreds}-\text{Tens}-\text{Ones}-[ . decimal][/tex]

The place value names before the decimal point in the number are

[tex][decimal.]\text{Tenths}-\text{Hundredths}-\text{Thousandths}-\text{Ten thousandths}-\cdots[/tex]

So, in the given number, the place value of

(a) Tens=2

(b) Ten thousandths=8

(c) Tenths=4

(d) Ones=1

Answer:

Step-by-step explanation:

Answer:D i think

Step-by-step explanation:

Answer:

15409.75

Step-by-step explanation

61630/ 4

=

30815/ 2

(Decimal: 15407.5)

Answer:

-2 | 7

0 | -1

3 | -13

-1 | 3

Step-by-step explanation:

i just did this question

Answer:

x = 3

Step-by-step explanation:

5.8x - 1.4 = 16                   {add 1.4 to both sides}

       5.8x = 16 + 1.4

       5.8x = 17.4              {Divide both sides by 5.8}

            x = 17.4/5.8

           x = 3

Answer:

i am pretty sure it is 7 and 8

Step-by-step explanation:

7 and 8 are before and after each other, so they are consecutive. 7 + 8 is 15

Answer:

it will be C

its 0.7 less than one

13 apples for 2 pies.

The unit rates.

13 apples = 2 pies

or, 1 pie = 13/2 = 6.5 apples

Option a. 6.5 apples per pie.

Answer:

21.45 mi

Step-by-step explanation:

Distance = speed × time

The time it takes the lion is:

13.75 mi = 50 mi/hr × t

t = 0.275 hr

The distance ran by the cheetah in that time is:

d = 78 mi/hr × 0.275 hr

d = 21.45 mi

Answer:

21.45 miles

Step-by-step explanation:

First, let's find the main ratio:

50:78

Now, let's bring a variable into the equation.

13.75:x

We can find the number to multiply by dividing 13.75 by 50 using long division.

13.75÷50=0.275

We need to find how many miles the cheetah ran, so we will multiply 78 and 0.275.

78·0.275=21.45

So, the cheetah ran 21.45 miles in the time it took the lion to run 13.75 miles.

Answer:

[tex]\frac{1}{729} c^{27}[/tex]

Step-by-step explanation:

[tex]\left(9c^{-9}\right)^{-3}=\frac{1}{\left(9c^{-9}\right)^3}[/tex]

[tex]\frac{1}{9^3\left(c^{-9}\right)^3}[/tex]

[tex]\frac{1}{729c^{-9\times 3}}[/tex]

[tex]\frac{1}{729c^{-27}}[/tex]

[tex]\frac{c^{27}}{729}=\frac{1}{729} c^{27}[/tex]

Answer:

w = 15.5

Step-by-step explanation:

31/2=15.5

Answer:

Step-by-step explanation:

If you meant w squared equals 31 (usually typed w^2 = 31), then to find w, take the square root.

[tex]w^2=31\\\\w=\sqrt{31} \approx 5.57[/tex]

Answer:

Answer: 5

Step-by-step explanation:

5(x+1)-5x

distribute

=(5)(x)+(5)(1)+ -5x

=5x+5+ -5x

Combine like terms

5x+5+ -5x

= (5x+ -5x) + 5

= 0 + 5

= 5

Answer:

(a) Sample Standard Deviation approximately to the nearest whole number = 6

(b) The use of Empirical Rule to make any general statements about the length of eruptions is empirical rules tell us about how normal a distribution and gives us an idea of what the final outcome about the length of eruptions is.

(c) The percentage of eruptions that last between 92 and 116 seconds using the empirical rule is 95%

(d) The actual percentage of eruptions that last between 92 and 116 seconds, inclusive is 95.45%

(e) The percentage of eruptions that last less than 98 seconds using the empirical rule is 16%

(f) The actual percentage of eruptions that last less than 98 seconds is 15.866%

Step-by-step explanation:

(a) Determine the sample standard deviation length of eruption.

Express your answer rounded to the nearest whole number.

Step 1

We find the Mean.

Mean = Sum of Terms/Number of Terms

= 90+ 90+ 92+94+ 95+99+99+100+100, 101+ 101+ 101+101+ 102+102+ 102+103+103+ 103+103+103+ 104+ 104+104+105+105+105+ 106+106+107+108+108+108 + 109+ 109+ 110+ 110+110+110+ 110+ 111+ 113+ 116+120/44

= 4582/44

= 104.1363636

Step 2

Sample Standard deviation = √(x - Mean)²/n - 1

=√( 90 - 104.1363636)²+ (90-104.1363636)² + (92 -104.1363636)² ..........)/44 - 1

= √(199.836777 + 199.836777 + 147.2913224+ 102.7458678+ 83.47314049+ 26.3822314+ 26.3822314+ 17.10950413+17.10950413+ 9.836776857+ 9.836776857, 9.836776857+9.836776857+ 4.564049585+ 4.564049585+ 4.564049585+ 1.291322313+ 1.291322313+ 1.291322313+ 1.291322313+ 1.291322313+ 0.01859504133+ 0.01859504133+ 0.01859504133+ 0.7458677685+ 0.7458677685+ 0.7458677685+ 3.473140497+ 3.473140497+ 8.200413225+ 14.92768595+ 14.92768595+ 14.92768595+ 23.65495868+ 23.65495868+ 34.38223141+ 34.38223141+34.38223141+ 34.38223141+ 34.38223141+47.10950414+ 78.56404959+ 140.7458677+ 251.6549586) /43

= √1679.181818/43

= √39.05073996

= 6.249059126

Approximately to the nearest whole number:

Mean = 104

Standard deviation = 6

(b) On the basis of the histogram drawn in Section 3.1, Problem 28, comment on the appropriateness of using the Empirical Rule to make any general statements about the length of eruptions.

The use of Empirical Rule to make any general statements about the length of eruptions is empirical rules tell us about how normal a distribution and gives us an idea of what the final outcome about the length of eruptions is .

(c) Use the Empirical Rule to determine the percentage of eruptions that last between 92 and 116 seconds.

The empirical rule formula states that:

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ .

2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ .

3)99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ

Mean = 104, Standard deviation = 6

For 68% μ - σ = 104 - 6 = 98, μ + σ = 104 + 6 = 110

For 95% μ – 2σ = 104 -2(6) = 104 - 12 = 92

μ + 2σ = 104 +2(6) = 104 + 12 = 116

Therefore, the percentage of eruptions that last between 92 and 116 seconds is 95%

(d) Determine the actual percentage of eruptions that last between 92 and 116 seconds, inclusive.

We solve for this using z score formula

The formula for calculating a z-score is is z = (x-μ)/σ

where x is the raw score, μ is the population mean, and σ is the population standard deviation.

Mean = 104, Standard deviation = 6

For x = 92

z = 92 - 104/6

= -2

Probability value from Z-Table:

P(x = 92) = P(z = -2) = 0.02275

For x = 116

z = 92 - 116/6

= 2

Probability value from Z-Table:

P(x = 116) = P(z = 2) = 0.97725

The actual percentage of eruptions that last between 92 and 116 seconds

= P(x = 116) - P(x = 92)

= 0.97725 - 0.02275

= 0.9545

Converting to percentage = 0.9545 × 100

= 95.45%

Therefore, the actual percentage of eruptions that last between 92 and 116 seconds, inclusive is 95.45%

(e) Use the Empirical Rule to determine the percentage of eruptions that last less than 98 seconds

The empirical rule formula:

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ .

2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ .

3)99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ

For 68% μ - σ = 104 - 6 = 98,

Therefore, 68% of eruptions that last for 98 seconds.

For less than 98 seconds which is the Left hand side of the distribution, it is calculated as

= 100 - 68/2

= 32/2

= 16%

Therefore, the percentage of eruptions that last less than 98 seconds is 16%

(f) Determine the actual percentage of eruptions that last less than 98 seconds.

The formula for calculating a z-score is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

For x = 98

Z score = x - μ/σ

= 98 - 104/6

= -1

Probability value from Z-Table:

P(x ≤ 98) = P(x < 98) = 0.15866

Converting to percentage =

0.15866 × 100

= 15.866%

Therefore, the actual percentage of eruptions that last less than 98 seconds is 15.866%

Answer:

the answer is B.

Step-by-step explanation:

30/5 is 6

Answer:

B)6

Step-by-step explanation:

the result after dividing

Answer:

it would be y = 9/2x

this is because you are trying to find the baking soda so you put that on the bottom, the sugar is just the multiplier for the baking soda

Step-by-step explanation:

Answer: 81/16

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