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lidiya [134]
3 years ago
8

2

Mathematics
1 answer:
GaryK [48]3 years ago
7 0

Answer:

B_2=4\pi r^2

see the explanation

Step-by-step explanation:

we know that

The area of the base of a cylinder is given by the formula

B=\pi r^{2}

where

r is the radius of the circular base

If the radius is doubled

then

r=2r

The new base area is

B_2=\pi (2r)^{2}

B_2=4\pi r^2

so

B_2=4B

therefore

The new area of the base is 4 times the area of the original base

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the difference between two numbers is 3.if the square of the smaller number is equal to four times the larger number, find the t
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Denote the numbers x, y. so we have equations y-x=3; x^2=4y, therefore x^2-4x-12=0. so x=6, y=9 or x=-2; y=1
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3 years ago
Determine whether the value given below is from a discrete or continuous data set. In a test of a method of gender selection, 72
Dmitriy789 [7]

Answer:

D. A discrete data set because there are a finite number of possible values.

Step-by-step explanation:

Assuming in a test of a method of gender selection, 725 couples used the XSORT method and 368 of them had baby girls. The value given is from a discrete data set because there are a finite number of possible values.

In Mathematics, a discrete data is a data set in which the number of possible values are either finite or countable.

On the other hand, a continuous data is a data set having infinitely many possible values and those values cannot be counted, meaning they are uncountable.

<em>Hence, if 725 couples used the XSORT method and 368 of them had baby girls; this is a discrete data because the values (725 and 368) are finite and can be counted. </em>

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3 years ago
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4 years ago
The art club had an election to select a president. 20% of the 75 members of the club voted in the election. How many members vo
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Step-by-step explanation:

6 0
3 years ago
A golfer would like to test the hypothesis that the variance of his golf score equals 10.0. A random sample of 28 rounds of golf
Ratling [72]

Answer:

Step-by-step explanation:

Hello!

Interest hypothesizes is that the variance of the golfer's scores equals to δ²= 10.0.

A random sample of 28 rounds of golf had a sample standard deviation S= 3.5

The statistics hypotheses are:

H₀: δ²= 10.0

H₁: δ²≠ 10.0

α: 0.05

To conduct a hypothesis test for the population variance you have to work using the Chi-Square distribution, this test is two-tailed so you will have two critical values. Between these two values is defined as the "not rejection region" and below and above them lies the "rejection region"

Lower critical value: X^2_{n-1; \alpha /2}= X^2_{27; 0.05}= 16.928

Upper critical value: X^2_{n-1;1-\alpha /2}= X^2_{27;0.95}= 40.113

If X^2_{H_0} ≤ 16.928 or X^2_{H_0} ≥ 40.113, the decision is to reject the null hypothesis.

If 16.928 < X^2_{H_0} < 40.113, the decision is to not reject the null hypothesis.

X^2= \frac{(n-1)S^2}{Sigma^2} ~~X^2_{n-1}

X^2_{H_0}= \frac{(28-1)12.25}{10} = 33.075

The calculated X^2_{H_0} is between the two critical values, so the decision is to not reject the null hypothesis. We can conclude that the population variance fro this golfers play score is 10.

I hope this helps!

5 0
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