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sergiy2304 [10]
3 years ago
10

What is the probability of selecting a seventh-grader from a school that has 250 sixth-graders, 225 seventh-graders, and 275 eig

hth-graders?
Mathematics
1 answer:
Nina [5.8K]3 years ago
3 0

Answer:

3/10 or 0.3 or 30%

Step-by-step explanation:

Its 225/750 total students

We simplify this by dividing by 5

45/150

Divide by 5 again

9/30

Divide by 3

3/10

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What is the opposite of the opposite of the number located at point a?
julsineya [31]

Answer:

A

Step-by-step explanation:

The opposite of point A is 9

Because the opposite of -9 is 9

And the opposite of 9 is -9

3 0
3 years ago
-3 1/2 - 2 1/3 explained
lakkis [162]
The answer is 5 5/6 simplified to the max.

rewrite the equation with separated parts

-3 1/2 - 2 1/3

-3 - 2 = -5

-1/2 - 1/3? u need to find the LEAST common denominator

-3/6 - 2/6 = -5/6

how did I get 6?

Rewriting input as fractions if necessary:
1/2, 1/3

For the denominators (2, 3) the least common multiple (LCM) is 6.

Therefore, the least common denominator (LCD) is 6.

Now lastly combine them total and its -5 - 5/6 = -5 5/6.


5 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
According to a survey of adults, 64 percent have money in a bank savings account. If we were to survey 50 randomly selected adul
zheka24 [161]

Answer:

The mean number of adults who would have bank savings accounts is 32.

Step-by-step explanation:

For each adult surveyed, there are only two possible outcomes. Either they have bank savings accounts, or they do not. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

E(X) = np

In this problem, we have that:

p = 0.64

If we were to survey 50 randomly selected adults, find the mean number of adults who would have bank savings accounts.

This is E(X) when n = 50.

So

E(X) = np = 50*0.64 = 32

The mean number of adults who would have bank savings accounts is 32.

6 0
3 years ago
A right triangle has a right exterior angle? <br> True or False?
adelina 88 [10]
False because 360-90 is 270
4 0
3 years ago
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