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3 years ago

Can someone help me please!

Mathematics

1 answer:

Hatshy [7]3 years ago

3 0

So we’re looking how many combinations we can make by selecting 5 items in a group of 17.

So we use n C r.

Where I the number of combinations of n objects taken r at a time.

So, 17 C 5

17! / (17-5)! 5! = 6188

(I hope this is correct)

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Which list shows the lengths from shortest to longest ?

wel

Answer:

It’s a

Step-by-step explanation:

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8 0

3 years ago

Measure of Angle 1 is 5x.<br> Measure of Angle 3 is x-24.<br><br> What is the measure of Angle 1?

olga55 [171]

Answer:

The answer to your question is: 30°

Step-by-step explanation:

Data

m∠ 1 = 5x

m∠ 3 = x - 24

Process

m ∠ 1 = m ∠ 3 because they are vertical angles

5x = x - 24

5x - x = 24

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x = 24 / 4

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4 0

3 years ago

12% of children are nearsighted, but this condition often is not detected until they go to kindergarten. A school district tests

Klio2033 [76]

Answer:

There is a 34.60% probability that 0 or 1 of them is nearsighted.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they are nearsighted, or they are not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem, we have that:

12% of children are nearsighted. This means that $p = 0.12$ .

A school district tests all incoming kindergarteners' vision. In a class of 18 kindergarten students, what is the probability that 0 or 1 of them is nearsighted?

There are 18 students, so $n = 18$