Mr martins math test which is worth 100 points has 29 problems. Each problem is worth either 5 points or 2 points.

dimulka [17.4K]

It would be 1 and 6/7

8 0

4 years ago

It is found that the number of raisins in a box of a popular cereal is normally distributed, with a mean of 133 raisins per box

gogolik [260]

Answer:

1.4

Step-by-step explanation:

We have been given that the number of raisins in a box of a popular cereal is normally distributed, with a mean of 133 raisins per box and a standard deviation of 10 raisins. My cereal box has 147 raisins.

To find the z-score for this box of cereal, we will use z-score formula.

$z=\frac{x-\mu}{\sigma}$ , where,

z = Z-score,

x = Sample score,

$\mu$ = Mean,

$\sigma$ = Standard deviation.

Upon substituting our given values, we will get:

$z=\frac{147-133}{10}$

$z=\frac{14}{10}$

$z=1.4$

Therefore, the z-score for this box of cereal is 1.4.

7 0

4 years ago

Sue Ann plans to apply waterproof coating to a parking pad that is 13 feet wide and 13 feet long. She buys a can of coating that

zheka24 [161]

a. she will not have enough coating

6 0

3 years ago

Read 2 more answers

A nonprofit organization sent mailing labels along with a request for donations to a random sample of 75,000 potential donors on

Law Incorporation [45]

Answer:

Check the explanation

Step-by-step explanation:

The null and alternative hypotheses in symbols and words are as follows:

: p ≤ 0.06 (There has been no change in the contribution rate due to addition of the mailing labels)

: p > 0.06 (The addition of the mailing labels has increased the contribution rate)

Test Statistics

The following information is provided: The sample size is N = 75000, the number of favorable cases is X = 5625, and the sample

The z-statistic is computed as follows:

- $P_0$ (1 -

Test statistic z = 17.297

P-value

P-value corresponding to z = 17.297 for a two tailed test is obtained using standard normal table.

p-value = 0.0000

90% Confidence Interval Calculation

Kindly check the attached image below.

Conclusion

Since p-value = 0.0000 < α = 0.05, we reject null hypothesis .

At 0.05 significance level, there is enough evidence to conclude that the addition of the mailing labels has increased the contribution rate.

Since the 90% confidence interval (0.073, 0.077) does not contain the hypothesized value 0.06, the result is significant and we reject null hypothesis.

In other words, at 0.10 significance level, there is enough evidence to conclude that the addition of the mailing labels has increased the contribution rate.