Solve for x. x2 = 16

A survey of an athletic shoe store’s customers showed that 8% of the customers run every day. The store has approximately 1900 c

dem82 [27]

Find 8% of 1900.

0.08 * 1900 = 152

So 152 customers run every day.

7 0

3 years ago

1:

bixtya [17]

Answer:

First day: 45x +20y = 875

Second day: 25x + 40y = 775

Adult ticket price: $15

Step-by-step explanation:

See paper attached. (:

8 0

3 years ago

Average precipitation for the first 7 months of the year, the average precipitation in toledo, ohio, is 19.32 inches. if the ave

Colt1911 [192]

Part A:

The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\sigma}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:

$P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\ \\ =1-P(z\ \textless \ -0.5410) \\ \\ =1-0.29426=\bold{0.7057}$

Part B:

</span>The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:

</span> $P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}$

7 0

3 years ago

John receives two credit card offers in the mail. Credit Card A: 1% introductory APR, 19.9% APR for purchases after the introduc

musickatia [10]

Answer:

(D) "...because he will not have to pay any interest on his purchases during the introductory APR period."

Step-by-step explanation:

There may be many reasons for John to choose one over the other depending on his future and planned usage patterns. However, answer the (D) is the only correct one in terms of stating a reason that is actually supported by the information in the question. Specifically, there is 0% APR for card B.

What's wrong with the other choices:

Choice A: incorrect statement re APR after intro period (is not 1%); B: incorrect statement re cash advance (is not 0%); C: statement is not supported by anything in the information given (it is an assumption, and is unlikely).

5 0

3 years ago

Read 2 more answers

I NEED HELP ON THIS SO CROWN FOR ANSWER

garri49 [273]

Answer:

Step-by-step explanation:

16oz is one pound and the puppy is already 514 so not it's

514.5(decimal)

$514\frac{1}{2}$ (fraction)

4 0

3 years ago