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Inessa [10]
3 years ago
10

What is the simplified result of following the steps below in order?

Mathematics
1 answer:
Keith_Richards [23]3 years ago
3 0
(5y+2x)*3+(x+y)=15y+6x-x-y=5x+14y
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A pooled variance is an estimated weighted variance made up of several different populations when the mean of each population ma
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Answer:

True

Step-by-step explanation:

Considering the above definition of Pooled Variance, the correct answer is TRUE.

This is because Pooled variance is used to determine the reasonable estimates of variance, where several repeated tests are expected at each value.

This helps to provide greater precision estimates of variance.

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If you answered 37 problems correctly on a 42 question test, what percent would you recieve
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A circle has a circumference of 16π.<br> What is the radius of the circle?
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3 years ago
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According to an​ article, 47​% of adults have experienced a breakup at least once during the last 10 years. Of 9 randomly select
In-s [12.5K]

Answer:

a) P(X=5)=(9C5)(0.47)^5 (1-0.47)^{9-5}=0.228

b) P(X=0)=(9C0)(0.47)^0 (1-0.47)^{9-0}=0.0033

And replacing we got:

P(X \geq 1)= 1-P(X

c) P(X=4)=(9C4)(0.47)^4 (1-0.47)^{9-4}=0.257

P(X=5)=(9C5)(0.47)^5 (1-0.47)^{9-5}=0.228

P(X=6)=(9C6)(0.47)^6 (1-0.47)^{9-6}=0.135

And adding we got:

P(4 \leq X \leq 6)= 0.257+0.228+0.135=0.620

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=9, p=0.47)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

Assuming the following questions:

a. exactly five

For this case we can use the probability mass function and we got:

P(X=5)=(9C5)(0.47)^5 (1-0.47)^{9-5}=0.228

b. at least one

For this case we want this probability:

P(X \geq 1)

And we can use the complement rule and we got:

P(X \geq 1)= 1-P(X

P(X=0)=(9C0)(0.47)^0 (1-0.47)^{9-0}=0.0033

And replacing we got:

P(X \geq 1)= 1-P(X

c. between four and six, inclusive.

For this case we want this probability:

P(4 \leq X \leq 6)

P(X=4)=(9C4)(0.47)^4 (1-0.47)^{9-4}=0.257

P(X=5)=(9C5)(0.47)^5 (1-0.47)^{9-5}=0.228

P(X=6)=(9C6)(0.47)^6 (1-0.47)^{9-6}=0.135

And adding we got:

P(4 \leq X \leq 6)= 0.257+0.228+0.135=0.620

6 0
3 years ago
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