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PilotLPTM [1.2K]
3 years ago
9

N(a)=60% n(o)=70% N(ano)=400 n(auo)complenment=10 find U and a only

Mathematics
1 answer:
Vlad1618 [11]3 years ago
5 0

n(A∪B)=n(A)+n(B)−n(A∪B)=50+60−40=70

n(AΔB)=n(A∪B)−n(A∩B)

⇒70−40=30.

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Find the value of the expression 4a to the fourth power - 2b squared +40 when a =2 and b=7
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. Exam scores for a large introductory statistics class follow an approximate normal distribution with an average score of 56 an
Andru [333]

Answer:

0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 56, \sigma = 5, n = 20, s = \frac{5}{\sqrt{20}} = 1.12

What is the probability that the average score of a random sample of 20 students exceeds 59.5?

This is 1 subtracted by the pvalue of Z when X = 59.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{59.5 - 56}{1.12}

Z = 3.1

Z = 3.1 has a pvalue of 0.9990.

So there is a 1-0.9990 = 0.001 = 0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

3 0
3 years ago
Find the equation of the circle with a diameter whose end points are (-1,-2) and (-3,2)
Sergeeva-Olga [200]

Answer:

The equation of the circle with a diameter whose end points are (-1,-2) and (-3,2) is

x^{2} +y^{2}+4x-1=0

Step-by-step explanation:

<u>Explanation:</u>-

<u>Step 1:</u>-

The equation of the circle having center and radius is

(x-h)^2+(y-k)^2=r^2

here center is (h,k) and radius is r

Given diameter whose end points are (-1,-2) and (-3,2)

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      (\frac{-1 +(-3) }{2} ,\frac{-2+2 }{2}  )

    (-2,0)

therefore center (h,k) = (-2,0)

<u>Step 2:-</u>

we have to find the radius of the circle

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r=\sqrt{{(-3)-(-1) ) ^{2}+ (2-(-2)) ^{2}}

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<u>Step 3</u>:-

center (h,k) = (-2,0) and

radius r=\sqrt{5}

The standard form of circle equation

(x-h)^2+(y-k)^2=r^2

(x-(-2))^2+(y-0)^2=\sqrt{5} ^2

on simplification is

x^{2} +y^{2}+4 x-1=0

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