C Find the product - ab and ac

Sladkaya [172]

Step-by-step explanation:

<h3>In Solving This You Need to Change It Into Improper Fractions First;</h3>

1 2/5×3 1/2

•You then multiply the 5 by the 1 and then add it to the 2 which is then going to be;

7/5× 3 1/2

•Then you multiply the 2 by the 3 and then add it to the 1;

7/5×7/2

•Then you proceed to multiply the first numerator by the second numerator and then also multiply the first denominator by the second denominator which will be;

14/10

After this you go on to simplify which is going to be;

7/5(NB: It was divided by two)

So mixed number in it's simplest form is going to be;

1 2/5

In Decimal form would have been; 1.4

Since the question is asking for mixed number in it's simplest form it's going to be;

1 2/5

8 0

3 years ago

Pls help its due today if it's correct I'll make you Brainliest!!!!

oee [108]

Answer:

b or c

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

PLZ HELP IT IS DUE TODAY AND WORTH 100 POINTS. PLEASE HELP I WILL BE SO GRATEFUL!!!! + BRAINLIEST

s344n2d4d5 [400]

Question 1snd 2 is not clear but this is the answers for 3,4,5 picture attached

3 0

3 years ago

The process that begins with recording business transactions and includes the completion of the financial statements is the

just olya [345]

Answer: "accounting cycle" .
_______________________________________________________

6 0

3 years ago

A machine that puts corn flakes into boxes is adjusted to put an average of 15.6 ounces into each box, with standard deviation o

Anastaziya [24]

Answer:

$\chi^2 =\frac{15-1}{0.0676} 0.1521 =31.5$

$p_v =P(\chi^2_{14} >31.5)=0.0047$

If we compare the p value and the significance level assumed of 0.05 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 higher than 0.26^2=0.0676.

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"

$n=15$ represent the sample size

$\alpha$ represent the confidence level

$s^2 =0.39^2 =0.1521$ represent the sample variance obtained

$\sigma^2_0 =0.26^2 =0.0676$ 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 violated, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \leq 0.0676$

Alternative hypothesis: $\sigma^2 >0.0676$

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.

$\chi^2 =\frac{15-1}{0.0676} 0.1521 =31.5$

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=15-1=14. And since is a right tailed test the p value would be given by:

$p_v =P(\chi^2_{14} >31.5)=0.0047$

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

"=1-CHISQ.DIST(31.5,14,TRUE)"

Conclusion

If we compare the p value and the significance level assumed of 0.05 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 higher than 0.26^2=0.0676.

4 0

3 years ago

CFind the product- ab and ac​

CFind the product- ab and ac