Please help xxxxxxxx

Sarah has 20 bags of chips for her party 1/4 of the chips were doritos and 1/5 of the remaining chips are sour cream and onion h

LuckyWell [14K]

Answer:
3 bags of sour cream and onion crisps

Explanation:
1/4 of 20 is 20 divided by 4 so there are 5 packs of Doritos.

20-5=15 packs still left

1/5 of the remaining bags is 1/5 of 15. 15 divided by 5 is 3.

There are 3 bags of sour cream and onion crisps.

Hope this helped!

5 0

3 years ago

Read 2 more answers

How do you turn 22% into a fraction in simplest form

sertanlavr [38]

Take 22 divided by 100. To get a decimal and then put that over 100 so the answer would be 22/100 I think I could be wrong

7 0

3 years ago

Read 2 more answers

The heights of 1/4 sheet cakes baked by a bakery have been normally distributed with a mean of μ = 2.05 inches and a standard de

jasenka [17]

Answer:

1. Null hypothesis: $\mu = 2.05$

Alternative hypothesis: $\mu \neq 2.05$

2. $P(Z>a)=0.025, P(Z$

And the value of a that satisfy this is a=1.96. So our critical regions are: $(-\infty,-1.96), (1.96,\infty)$

3. $z=\frac{2.01-2.05}{\frac{0.12}{\sqrt{9}}}=-1$

4. If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean for the heights is NOT significant different than 2.05. Using the critical region founded on part 2 we agree with the decision obtained with the p value since -1 is not on the critical zones, so we FAIL to reject the null hypothesis.

Step-by-step explanation:

Data given and notation

$\bar X=2.01$ represent the sample mean

$\sigma=0.12$ represent the standard deviation for the population

$n=9$ sample size

$\mu_o =2.05$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

1) The null and alternative hypotheses should be ?

We need to conduct a hypothesis in order to determine if the mean change from 2.05, the system of hypothesis would be:

Null hypothesis: $\mu = 2.05$

Alternative hypothesis: $\mu \neq 2.05$

2. Using the critical value to set up the decision rule, the decision rule should be?

For this case we need two critical values since we are conducting a two tailed test. We have this equality:

$P(Z>a)=0.025, P(Z$

And the value of a that satisfy this is a=1.96. So our critical regions are: $(-\infty,-1.96), (1.96,\infty)$

3. The test statistic of this test is?

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$z=\frac{2.01-2.05}{\frac{0.12}{\sqrt{9}}}=-1$

4. The conclusion of this test should be

Since is a two tailed test the p value would be:

$p_v =2*P(Z$

Conclusion

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean for the heights is NOT significant different than 2.05. Using the critical region founded on part 2 we agree with the decision obtained with the p value since -1 is not on the critical zones, so we FAIL to reject the null hypothesis.

4 0

3 years ago

Suppose there were 4,065,014 births in a given year. Of those births 2,081,287 were boys and 1,983,727 were girls. What is the p

kogti [31]

Answer:

The required probability is $\simeq 0.26214$