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S_A_V [24]
2 years ago
12

The circumference of a circle is equal to n times its diameter. Estimate the circumference of a round swimming pool that has a d

iameter of 24 feet. Use n=3.14. *
Mathematics
1 answer:
sdas [7]2 years ago
5 0

Answer:  An estimate done quickly: The circumference is about 75 ft.

Step-by-step explanation:

Round π to 3, and 24 to 25.  3 times 25 is 75

A more exact value would be 75.36

3.14 × 24 = 75.36

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Q3) f(x) 2x^2 - 5x, g(x) = 3x^3 find g(4x)
mrs_skeptik [129]

Answer:

g(4x) = 192x^3

Step-by-step explanation:

For this problem, f(x) is irrelevant since we are simply are dealing with g(x).  We will simply replace the value of x in g(x) with 4x.  So let's do that.

g(x) = 3x^3

g(4x) = 3(4x)^3

g(4x) = 3(4^3)(x^3)

g(4x) = 3(64)(x^3)

g(4x) = 192x^3

Hence, g(4x) is 192x^3.

Cheers.

4 0
3 years ago
How long should an escalator be if it is to make an angle 30 with the floor and
Charra [1.4K]

Answer:

it should be 23 feet long

Step-by-step explanation:

7 0
2 years ago
Find the selling price of a $36 item after a 50% markup
agasfer [191]

Answer:

$54

Step-by-step explanation:

36 divided by 2= 18

$36+$18=$54

7 0
2 years ago
Represent 8 as an expression.
klio [65]

Answer:

x + 8

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
The standard deviation of math test scores at one high school is 16.1. A teacher claims that the standard deviation of the girls
Elanso [62]

Answer:

\chi^2 =\frac{22-1}{259.21} 166.41 =13.482

p_v =P(\chi^2

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(13.482,21,TRUE)"

If we compare the p value and the significance level provided we see that p_v >\alpha so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population deviation is not significantly lower than 16.1 at 1% of significance.

Step-by-step explanation:

Notation and previous concepts

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

n=22 represent the sample size

\alpha=0.01 represent the confidence level  

s^2 =12.9^2=166.41 represent the sample variance obtained

\sigma^2_0 =16.1^2 =259.21 represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population deviation is smaller than 16.1 (that's equivalent to check if the population variance is lower than 259.21:

Null Hypothesis: \sigma^2 \geq 259.21

Alternative hypothesis: \sigma^2

Calculate the statistic  

For this test we can use the following statistic:

\chi^2 =\frac{n-1}{\sigma^2_0} s^2

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

\chi^2 =\frac{22-1}{259.21} 166.41 =13.482

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 21. And since is a left tailed test the p value would be given by:

p_v =P(\chi^2

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(13.482,21,TRUE)"

Conclusion

If we compare the p value and the significance level provided we see that p_v >\alpha so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population deviation is not significantly lower than 16.1 at 1% of significance.

3 0
3 years ago
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