All categories

2 years ago

Hep I’m trash at math

Mathematics

2 answers:

sergejj [24]2 years ago

6 0

Sameeeeeeeeeeeeeeeeeee

puteri [66]2 years ago

5 0

Step-by-step explanation:

$\pi {r}^{2} \frac{h}{3}$

π×4×4×8/3

π×16×8/3

128/3×π

128/3π cm²

You might be interested in

What am I suppose to do here? What is High range and Low range?

VLD [36.1K]

That means what is the least u would spend and most so like if someone asked you how much you would pay for a product, you may say $7-$10 and 7 would be the low range and 10
would be the high range

7 0

1 year ago

Read 2 more answers

5 1 point 34 is 68% of what number?

Ilya [14]

Answer:

Step-by-step explanation:

Fastest method for calculating 34 is 68 percent of what number. Assume the unknown value is 'Y' 34 = 68% x Y. 34 = 68 / 100 x Y Multiplying both sides by 100 and dividing both sides of the equation by 68 we will arrive at: Y = 3 x 100 / 68. Y = 50%. Answer: 34 is 68 percent of 50

4 0

2 years ago

What is the equation for the black line?

BARSIC [14]

-3 and 5..................

8 0

2 years ago

A marketing research firm wishes to compare the prices charged by two supermarket chains—Miller’s and Albert’s. The research fir

AVprozaik [17]

Answer:

a) $F=\frac{s^2_2}{s^2_1}=\frac{1.84^2}{1.4^2}=1.727 \approx 1.73$

b) For this case since we have a two tailed test we have two critical values, and we can find the two values since we need that on each tail the area would be $\alpha/2 =0.025$ , the distribution is the F with 9 df for the numerator and denominator, and we can use the following excel codes to find the critical values:

"=F.INV(0.025,9,9)"

"=F.INV(1-0.025,9,9)"

The two critical values are 0.248 and 4.026, so the rejection zone would be: $F>4.026 \cup F$ since our calculated value is not on the rejection zone we fail to rejec the null hypothesis

Step-by-step explanation:

Data given and notation

$n_1 = 10$ represent the sampe size for the Miller's Stores

$n_2 =10$ represent the sample size for the Albert's stores

$\bar X_1 =121.92$ represent the sample mean for Miller's store

$\bar X_2 =114.81$ represent the sample mean for Albert's store

$s_1 = 1.4$ represent the sample deviation for the Miller's store

$s_2 = 1.84$ represent the sample deviation for the Albert's stores

$s^2_2 = 12.25$ represent the sample variance for the utility stocks

$\alpha=0.05$ represent the significance level provided

Confidence =0.95 or 95%

F test is a statistical test that uses a F Statistic to compare two population variances, with the sample deviations s1 and s2. The F statistic is always positive number since the variance it's always higher than 0. The statistic is given by:

$F=\frac{s^2_2}{s^2_1}$

Solution to the problem

System of hypothesis

We want to test if the variation for th two groups is the same, so the system of hypothesis are:

H0: $\sigma^2_1 = \sigma^2_2$

H1: $\sigma^2_1 \new \sigma^2_2$

a) Calculate the statistic

Now we can calculate the statistic like this:

$F=\frac{s^2_2}{s^2_1}=\frac{1.84^2}{1.4^2}=1.727 \approx 1.73$

Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have $n_1 -1 =10-1=9$ and for the denominator we have $n_2 -1 =10-1=9$ and the F statistic have 9 degrees of freedom for the numerator and 9 for the denominator. And the P value is given by:

P value

$p_v =2*P(F_{9,9}>1.727)=0.428$

And we can use the following excel code to find the p value:"=2*(1-F.DIST(1.727,9,9,TRUE))"

b) Critical value

For this case since we have a two tailed test we have two critical values, and we can find the two values since we need that on each tail the area would be $\alpha/2 =0.025$ , the distribution is the F with 9 df for the numerator and denominator, and we can use the following excel codes to find the critical values:

"=F.INV(0.025,9,9)"

"=F.INV(1-0.025,9,9)"

The two critical values are 0.248 and 4.026, so the rejection zone would be: $F>4.026 \cup F$ since our calculated value is not on the rejection zone we fail to rejec the null hypothesis

Since our calcu

Conclusion

Since the $p_v > \alpha$ we have enough evidence to FAIL to reject the null hypothesis. And we can say that we don't have enough evidence to conclude that the two deviations are different at 5% of significance.