2/3 divided by 9/10 simplified step-by-step

Let’s assume the 147 freshman maintained a 65% attendance rate during the first 4 week of school. If our next 4 week goal for at

shutvik [7]

Answer:

12 students

Step-by-step explanation:

If 65% equals 147

Then to maintain 70% there is a need for 5% more

<u>Which equals</u>

5*147/65 ≈ 12 students

4 0

3 years ago

Read 2 more answers

PLEASE HELP. Prove that the value of the expression: b (36^5−6^9)(38^9−38^8) is divisible by 30 and 37.

pishuonlain [190]

Answer:

Step-by-step explanation:

First factor the second binomial

38^9 - 38^8 = 38^9 (38 - 1 ) = 38^9(37)

So there's your 37. This whole expression is divisible by 37

Now do the first binomial

36^5 - (36^4)*6

36^4(36 - 6)

36^4(30) and there's your thirty.

That first term is going to cause a bit of trouble showing that 36^4 * 6 = 6^9

6(36^4) = 6(6^2)^4 = 6^1 * 6^8 = 6^9

So this factors into 30*(36^4)*37*38^9

3 0

4 years ago

The weights of 67 randomly selected axles were found to have a variance of 3.85. Construct the 80% confidence interval for the p

VLD [36.1K]

Answer:

$3.13$

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 67

Variance = 3.85

We have to find 80% confidence interval for the population variance of the weights.

Degree of freedom = 67 - 1 = 66

Level of significance = 0.2

Chi square critical value for lower tail =

$\chi^2_{1-\frac{\alpha}{2}}= 51.770$

Chi square critical value for upper tail =

$\chi^2_{\frac{\alpha}{2}}= 81.085$

80% confidence interval:

$\dfrac{(n-1)S^2}{\chi^2_{\frac{\alpha}{2}}} < \sigma^2 < \dfrac{(n-1)S^2}{\chi^2_{1-\frac{\alpha}{2}}}$

Putting values, we get,

$=\dfrac{(67-1)3.85}{81.085} < \sigma^2 < \dfrac{(67-1)3.85}{51.770}\\\\=3.13$

Thus, (3.13,4.91) is the required 80% confidence interval for the population variance of the weights.

8 0

3 years ago

Suppose a local manufacturing company claims their production line has a variance of less than 9.0. A quality control engineer d

Alexxx [7]

Answer:

$\chi^2 =\frac{35-1}{9} 2.12^2 =16.979$

$p_v =P(\chi^2$

If we compare the p value and the significance level provided we see that $p_v$ so on this case we have enough evidence in order to reject the null hypothesis at the significance level provided. And that means that the population variance is significantly lower than 9

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"

$n=35$ represent the sample size

$\alpha=0.01$ represent the confidence level

$s =2.12$ represent the sample deviation obtained

$\sigma_0 =3$ represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population variance specification is lower than 9 and the deviation lower than 3, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \geq 9$

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.

$\chi^2 =\frac{35-1}{9} 2.12^2 =16.979$

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case df= n-1= 35-1=34. 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(16.979,34,TRUE)"

Conclusion

If we compare the p value and the significance level provided we see that $p_v$ so on this case we have enough evidence in order to reject the null hypothesis at the significance level provided. And that means that the population variance is significantly lower than 9

6 0

3 years ago

Which graph shows the points (−112, 12) and (1.5, −0.5) plotted correctly? On a coordinate plane, a point is 0.5 units to the ri

chubhunter [2.5K]

Answer:

Graph 'B' represents the correct graph.

The correct graph is attached below.

Step-by-step explanation:

Let 'A' be the point: