Mira has breakfast at a restaurant. She leaves a 20% tip of $1.80.

Two business partners, Alan and Rob, invested money in their business at a ratio of 6 to 7. Bob invested the greater amount. The

Verdich [7]

Answer:

180 Alan, 210 Rob

Step-by-step explanation:

6+7= 13

390÷13= 30

30×6=180 =Alan

30××7=210 =Rob

4 0

3 years ago

There are 6 time zones in the United States. The eastern part of the U.S., including New York City, is in the Eastern Time Zone.

VashaNatasha [74]

It’s morning in California

6 0

3 years ago

Will is w years old <br> Ben is 3 years older <br> Write an expression, in terms of w, for bens age

Westkost [7]

Ben is w+3 years old.

7 0

3 years ago

A yogurt company claims that it prints a free yogurt coupon under a randomly selected 20% of its lids. A loyal customer purchase

Alona [7]

Answer:

The customer can conclude that the company's claim is correct

Step-by-step explanation:

The percentage of lids that has a free yogurt coupon = 20%

The number of cups a loyal customer purchases = 85 yogurt cups

The number of cups that contained a coupon = 12 (14.1%)

The confidence interval performed = 99% confidence interval for the proportion of yogurt cups containing coupon codes

The interval obtained = (0.044, 0.238)

Therefore, the range of proportion within which the true proportion exists is 0.044 < $\hat p$ < 0.238

The range of percentage within which the true percentage exist is therefore;

0.044 × 100 = 4.4% < $\hat p$ × 100 < 0.238 × 100 = 23.8%

Given that the possible true percentage of lids that has a coupon is between 4.4% and 23.8% at 99% confidence level, the customer can conclude that only 12 of his yogurt cup contained coupon by chance and that the company's claim is correct.

4 0

3 years ago

f the dean wanted to estimate the proportion of all students receiving financial aid to within 3% with 99% reliability, how many

Oksanka [162]

Answer:

n=1849

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by: