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NeX [460]
3 years ago
12

A sprinkler head is equidistant from flower garden A and a small shrub B. The sprinkler waters in a

Mathematics
1 answer:
Natalija [7]3 years ago
7 0

Answer:

69^{\circ}

Step-by-step explanation:

We are given that

Length of minor arc AB=12 feet

Radius of circle=r=10 feet

Let x  be the measure of central angle subtended by minor arc AB.

We know that

Length of an arc=\frac{Central\;angle}{360}\times 2\pi r

Where \pi=3.14

Using the formula

12=\frac{x}{360}\times 2\times 3.14\times 10

x=\frac{12\times 360}{2\times 3.14\times 10}

x=69^{\circ}

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Answer:

3.438

Step-by-step explanation:

5.73 × 0.6 = 3.438

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Step-by-step explanation:

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Which relationship is always true for the angles RXY and Z or the triangle ABC
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Answer:

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Step-by-step explanation:

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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

This probability is:

P = P(X = 0) + P(X = 1)

So

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

P(X = 0) = C_{18,0}.(0.12)^{0}.(0.88)^{18} = 0.1002

P(X = 1) = C_{18,1}.(0.12)^{1}.(0.88)^{17} = 0.2458

So

P = P(X = 0) + P(X = 1) = 0.1002 + 0.2458 = 0.3460

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

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