What is the ratio of 24 packs 2 for $9

A doctor told a patient to drink 1.6 liters of water everyday. About how many liters should the patient drink in 14 days?

inn [45]

He would have drank 22.4 litres of water in 14 days

7 0

2 years ago

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The owner of a bakery wants to know whether customers have any preference among 5 different flavors of muffin. A sample of 200 c

Troyanec [42]

So,, there are 5 different flavors. A total of 180 people were asked. Hence, the hypothesis that there is no significant difference is that every flavor gets 180/5=36 flavors. x^2= $\sum{\frac{(x_i)^2}{m_i}}-n$ . In this case, mi is the proportion of the hypothesis, thus 36, n=180 and xi is the number of actual observations. Substituting the known quantities, we get that x^2=9. The degree of association is given by $\sqrt{X^2}/ \sqrt{N(n-1)}$ . This yields around 0.10, much higher than our limit.

8 0

3 years ago

Step 1 of 3: Each year a nationally recognized publication conducts its Survey of America’s Best Graduate and Professional Schoo

lukranit [14]

Answer:

Step-by-step explanation:

Hello!

Given the variables

Y: Starting salary of a graduate at a top business school

X: GMAT score of the business school

The sample used had a size n=20

The estimated regression equation was ^Y= -92040 + 228Xi

The correlation coefficients were r= 0.81

You need to know if r is statistically significant at a 0.05 level:

Assuming the hypotheses are:

H₀: There is a linear correlation between the variables of interest

H₁: There is no linear correlation between the variables of interest

α: 0.05

This test is two-tailed.

The p-value for a r=0.81 and n=20 is 0.000015

Using the p-value approach, the decision rule is:

If p-value ≤ α, reject the null hypothesis

If p-value > α, do not reject the null hypothesis

In this case, the p-value is less than α, the decision is to reject the null hypothesis.

At a 5% level, you can conclude that the coefficient of correlation is statistically significant.

4 0

3 years ago

in 1991, a gallup poll reported this percent to be 79%. using the data from this poll, test the claim that the percent of driver

snow_lady [41]

We reject our null hypothesis, H₀, at a level of significance of =0.01 since the P-value is less than that threshold. There is compelling statistical data to indicate that since 1991, the proportion of drivers who love driving has decreased.

Given,

The Pew Research Center recently polled n=1048 U.S. drivers and found that 69% enjoyed driving their automobiles.

In 1991, a Gallup poll reported this percentage to be 79%. using the data from this poll, test the claim that the percentage of drivers who enjoy driving their cars has declined since 1991.

To report the large-sample z statistic and its p-value,

Null hypothesis,

H₀ : p = 0.79

Alternative hypothesis,

Ha : p < 0.79

Level of significance, α = 0.01

Under H₀

Test statistic,

$Z_0= \frac{\hat p - p}{\sqrt{\frac{p(1-p)}{n} } }$

Z₀ = -7.948

The alternative hypothesis(Ha) is left-tailed, so the P-value of the test is given by

P-value = P(z <-7.945)

= 0.000 (from z-table)

Since the P-value is smaller than given level of significance, α=0.01 we reject our null hypothesis, H₀, at α=0.0.1 level Strong statistical evidence to conclude that the percentage of drivers who enjoy driving their cars has declined since 1991.

To learn more about hypothesis click here:

brainly.com/question/17173491

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5 0

1 year ago

Help me on this one please

shepuryov [24]

Answer:

The equation is set up wrong

Step-by-step explanation:

$9^{2} +b^{2} =15^{2}$

3 0

2 years ago