All categories

3 years ago

Each prize for a carnival booth costs $0.32. How many prizes can you buy with $96?

Mathematics

2 answers:

Dafna11 [192]3 years ago

7 0

Step one: divide 96 by .32.
Step two: get 300.
Step three: 300 is the answer

Dima020 [189]3 years ago

4 0

You can buy 300 toys

You might be interested in

If a = b + c and c ___ 0, then a > b .

Aleksandr-060686 [28]

The correct answer would be =

8 0

3 years ago

Read 2 more answers

Which statement is true about the ranges for the box plots?A number line goes from 0 to 16. Morning's whiskers range from 3 to 1

Mnenie [13.5K]

Answer:

The correct answer is A. The range for the Morning box plot is the same as the range for the Afternoon box plot.

I just took the test and this was the right answer

8 0

3 years ago

Read 2 more answers

Distributive property 22multiplyby5

Marizza181 [45]

110 is the answer of this question

7 0

2 years ago

Read 2 more answers

Kevin went to see his favorite sports team play his car can go 320 miles on one lank of gas. If he used 20% of a full tank of ga

Oxana [17]

Answer:

64 miles

Step-by-step explanation:

If he used 20% of the gas, he can go 20% of the miles

20% of 320 miles

.20 * 320

64 miles

7 0

3 years ago

Read 2 more answers

A fast food restaurant executive wishes to know how many fast food meals teenagers eat each week. They want to construct a 99% c

Lana71 [14]

Answer:

The minimum sample size required to create the specified confidence interval is 2229.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.005 = 0.995$ , so $z = 2.575$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

What is the minimum sample size required to create the specified confidence interval

This is n when $M = 0.06, \sigma = 1.1$

$0.06 = 2.575*\frac{1.1}{\sqrt{n}}$

$0.06\sqrt{n} = 2.575*1.1$

$\sqrt{n} = \frac{2.575*1.1}{0.06}$

$(\sqrt{n})^{2} = (\frac{2.575*1.1}{0.06})^{2}$

$n = 2228.6$

Rounding up

The minimum sample size required to create the specified confidence interval is 2229.

7 0

3 years ago