C=$7.00

C stands for the cost, and n would be the amount of hamburgers. Since we are looking to determine the costs of 14 hamburgers we would have to then multiply 0.5 and 14.

14 x 0.5 =7

So your answer is 7 dollars

7 0

3 years ago

Read 2 more answers

Can someone help me with 8 or 9

kumpel [21]

8
A. Delivery is $20. One cord is $120.
B. 1 1/2 cord costs $200. ($180 for the cord and $20 for the delivery)

(I can’t see the full problem for 9)

4 0

3 years ago

A random sample of 5120 permanent dwellings on an entire reservation showed that 1627 were traditional hogans.

krek1111 [17]

Answer:

a) $\hat p=\frac{1627}{5120}=0.3178$

b)The 99% confidence interval would be given by (0.301;0.355)

Step-by-step explanation:

1) Notation and definitions

$X=1627$ number of permanent dwellings on the entire reservation that are traditional hogans.

$n=5120$ random sample taken

$\hat p=\frac{1627}{5120}=0.318$ estimated proportion of permanent dwellings on the entire reservation that are traditional hogans.

$p$ true population proportion of permanent dwellings on the entire reservation that are traditional hogans.

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

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

Part a

The point of estimate for p=the proportion of all permanent dwellings on the entire reservation that are traditional hogans is given by:

$\hat p=\frac{1627}{5120}=0.3178$

Part b

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: