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notka56 [123]
3 years ago
11

If the width of a pool is 60 feet and the perimeter is 200 feet, how long is the pool?

Mathematics
2 answers:
velikii [3]3 years ago
8 0

Answer:

140

Step-by-step explanation:

200-60 = 140

denis-greek [22]3 years ago
5 0

Answer:

c

Step-by-step explanation:

Substitute the values in the problem to get, (200) = 2(60) + 2W. Solve for W to get 200 = 120 + 2W. Subtract 120 on both sides to get 80 = 2W. Divide by 2 on both sides to get 40 = W or a width of 40 feet.

hope that helped you.

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mestny [16]

Hello from MrBillDoesMath!

Answer:

8 (3w^2 - 2)


Discussion:

The  greatest common factor is 8, so

24w^2 - 16 = 8 (3w^2 - 2)


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MrB

6 0
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I will give brainliest<br> 7k+8=2k-37<br> Show work pleaseeeee
Artyom0805 [142]

Answer:

k = -9

Step-by-step explanation:

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Multiply each equation by a constant that would help to eliminate the y terms. 2x-5y=-21 3x-3y=-18 What are the resulting equati
anyanavicka [17]

Answer:

6x - 15y = -63

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Step-by-step explanation:

We need to multiply the two equations by a constant that will eliminate the y terms.

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2x - 5y = -21

3x - 3y = -18

Let us multiply the first by 3 and the second by 6. The resulting equations will be:

6x - 15y = -63

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8 0
3 years ago
A certain medical test is known to detect 73% of the people who are afflicted with the disease Y. If 10 people with the disease
Setler [38]

The probability of an event is the measurement of the chance of that event's occurrence. The probabilities of considered events are:

  • P(At least 8 have the disease) ≈ 0.4378
  • P(At most 4 have the disease)  ≈ 0.0342
<h3 /><h3>How to find that a given condition can be modeled by binomial distribution?</h3>

Binomial distributions consist of n independent Bernoulli trials. Bernoulli trials are those trials that end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))

Suppose we have random variable X pertaining to a binomial distribution with parameters n and p, then it is written as

X = B(n,p)

The probability that out of n trials, there'd be x successes is given by

P(X=x)  = ^nC_xp^x(1-p)^{n-x}

Since 10 people can be either diseased or not and they be so independent of each other (assuming them to be selected randomly) , thus, we can take them being diseased or not as outputs of 10 independent Bernoulli trials.

Let we say

Success= Probability of a diseased person tagged as diseased by the clinic

Failure = Probability of a diseased person tagged as not diseased by the clinic.

Then,

P(Success) = p = 72% = 0.72 (of a single person)

P(Failure) = q = 1-p = 0.28

Let X be the number of people diagnosed diseased by the clinic out of 10 diseased people. Then we have: X ≈ B(n+10,P=0.73)

Calculating the needed probabilities, we get:

a) P(At leased 8 have disease) = P(X≥8) =P(X=8) + P(X=9) + P(X=10)

P(X≥8) = ^{10}C_8(0.73)^8(0.28)^2+^{10}C_9(0.73)^9(0.27)^1+^{10}C_{10}(0.73)^{10}(0.27)^0

P(X≥8) ≈ 0.2548 + 0.1456 + 0.0374 ≈ 0.4378

b)  P(At most 4 have the disease) = P(X≤4) = P(X=0) + P(X=1)+P(X=2)+P(X=3)+P(X=4)

P (X ≤ 4) =

^{10}C_0(0.73)^0(0.27)^{10}+^{10}{C_1(0.73)^1(0.27)^9+^{10}{C_2(0.73)^2(0.27)^8+^{10}C_3(0.73)&^3(0.27)^7 \\

+^{10}C_4(0.73)^4(0.27)^6

P (X ≤ 4) = 0.000003 + 0.000076+0.00088+0.00604+0.02719

P (X ≤ 4) =  0.0342

Thus,

The probabilities of considered events are:

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  • P(At most 4 have the disease)  = 0.0342 approx

Learn more about binomial distribution here:

brainly.com/question/13609688

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4 0
2 years ago
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