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3 years ago

What is 2 to the power of 3 over 2 equal to? (5 points) squre root of 8 cube root of 8 cube root of 16 square root of 16

Mathematics

1 answer:

Airida [17]3 years ago

3 0

Answer:

Square root of 8.

Step-by-step explanation:

Given:

Number given : 2 to the power of 3 over 2 equal ...

Using the laws of exponent:

We know that:

⇒ $\sqrt{a} = (a)^1^/^2$ and

⇒ $\sqrt[3]{a}=(a)^1^/^3$

So,

According to the question:

2 to the power of 3 over 2 = $(2) ^3^/^2$

Using fractional exponent concept where : $\sqrt[y]{a^x} = (a)^x^/^y$

This can be re-written as:

⇒ $(2^3)^1^/^2$ and $\sqrt{2^3}$ that is equivalent to $\sqrt{8}$ as ... $2^3=2\times 2\times 2=8$

Square root of 8 is our final answer.

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The eatery has 200 tables there were reservations for 1/10 of tables how many tables were reserved

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If you see 1/10 of it means part of it and the answer will be less.
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3 years ago

Nine more than the quotient of 2 and a number of x

Naddika [18.5K]

Answer: 9 + 2/x - x = -19

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3 years ago

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son4ous [18]

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3 years ago

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Soloha48 [4]

Answers:

1a) y=-10/3x+90

1b) 20

1c) -18

2a) 2.8

2b) How much the heights of five basketball players vary from the average height.

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Explanations:

1a) The trend line is linear, so we just need to find the slope and y-intercept to find an equation for it. Our y-intercept is (0,90), or 90, and our slope is -10/3. Our equation is now y=-10/3x+90.

1b) To find when x=21, we plug 21 into our equation where the x is. Now we do the math.

y=-10/3(21)+90 (plug in)

y=-70+90 (simplify by multiplying -10/3 by 21)

y=20 (simplify by adding -70 to 90)

Therefore, we can predict that when x is 21, y is 20.

1c) To find when y=150, we plug 150 into our equation where the y is. Now we do some more math.

150=-10/3x+90 (plug in)

60=-10/3x (subtract 90 from both sides

-18=x (divide both sides by -10/3)

Therefore, we can predict that when y is 150, x is -18.

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2a) The mean absolute deviation (or MAD for short) of a data set is calculated as such:

Step 1) Find the mean (average) by finding the sum of the data values, then dividing the sum of the data values by the number of data values. In this case, we have the numbers 65, 58, 64, 61, and 67, which add up to 315. The data set has 5 numbers, so we divide our sum of 315 by 5 to get 63. Therefore, our mean is 63.

Step 2) Find the absolute value of the distance between each data value and the mean. In this case, we find out how far away each data value is from 63, our mean. To do this, we subtract 63 from each number.

65-63=2

58-63=-5

64-63=1

61-63=-2

67-63=4

Some of these values are negative, but we're using absolute value so they all become positive. We now have a new set of values: 2, 5, 1, 2, and 4.

Step 3) Finally, we calculate the mean of our new set of values. In this case, we will add up 2, 5, 1, 2, and 4 to get 14 and divide by 5 to get our MAD of 2.8. Therefore, the MAD (and the answer to problem 2a) is 2.8.

2b) Now we just find out what the MAD means in this context. The MAD always is a measure of variance in a data set. In this context, it's describing how much the heights (in inches) of five people on a basketball team vary from the average height.

Hope this helps!

3 0

3 years ago

Listed below are the number of years it took for a random sample of college students to earn bachelor's degrees (based on data f

schepotkina [342]

Answer:

a) Mean = 6.5, sample standard deviation = 3.50

b) Standard error = 0.7826

c) Point estimate = 6.5

d) Confidence interval: (5.1469 ,7.8531)

Step-by-step explanation:

We are given the following data set for students to earn bachelor's degrees.

4, 4, 4, 4, 4, 4, 4.5, 4.5, 4.5, 4.5, 4.5, 4.5, 6, 6, 8, 9, 9, 13, 13, 15

a) Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{130}{20} = 6.5$

Sum of squares of differences = 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 4 + 4 + 4 + 4 + 4 + 4 + 0.25 + 0.25 + 2.25 + 6.25 + 6.25 + 42.25 + 42.25 + 72.25 = 233.5

$S.D = \sqrt{\frac{233.5}{19}} = 3.50$

b) Standard Error

$= \displaystyle\frac{s}{\sqrt{n}} = \frac{3.50}{\sqrt{20}} = 0.7826$

c) Point estimate for the mean time required for all college is given by the sample mean.

$\bar{x} = 6.5$

d) 90% Confidence interval:

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at degree of freedom 19 and}~\alpha_{0.10} = \pm 1.729$

$6.5 \pm 1.729(\frac{3.50}{\sqrt{20}} ) = 6.5 \pm 1.3531 = (5.1469 ,7.8531)$

e) No, the confidence interval does not contain the value of 4 years. Thus, confidence interval is not a good estimator as most of the value in the sample is of 4 years. Most of the sample does not lie in the given confidence interval.

5 0

4 years ago