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3 years ago

7

Jim's father is older than 40 but younger than 50. If you divide his age by 2, 4, 5, 8, or 10, there will be a remainder of 1, H

ow old is Jim's father?

1 answer:

ExtremeBDS [4]3 years ago

8 0

Answer:

41

Step-by-step explanation:

41:2= 20 with remainder of 1

41:4=10 r:1

41:5=8 r:1

41:8=5 r:1

41:10=4 r:1

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Harlamova29_29 [7]

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0.7995 = 79.95% probability that the sample will contain at least three defectives.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Suppose that a random sample of 20 items is selected from the machine.

This means that $n = 20$

The machine produces 20% defectives

This means that $p = 0.2$

Mean and standard deviation:

$\mu = E(X) = np = 20*0.2 = 4$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{20*0.2*0.8} = 1.79$

Probability that the sample will contain at least three defectives

Using continuity correction, this is $P(X \geq 3 - 0.5) = P(X \geq 2.5)$ , which is 1 subtracted by the pvalue of Z when X = 2.5. So

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

$Z = \frac{2.5 - 4}{1.79}$

$Z = -0.84$

$Z = -0.84$ has a pvalue of 0.2005

1 - 0.2005 = 0.7995

0.7995 = 79.95% probability that the sample will contain at least three defectives.

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3 years ago

The box plots summarize the data for heights of players on a national

andreev551 [17]

(Diagram of the box plots is attached below)

Options:

A. 75% of the volleyball players’ heights are equal to or greater than the median basketball players’ heights

B. 75% of the volleyball players’ heights are equal to or greater than the median basketball players’ heights

C. 25% of the basketball players’ heights are equal to or greater than the median volleyball players’ heights

D. 75% of the basketball players’ heights are equal to or greater than the median volleyball players’ heights

Answer:

D. 75% of the basketball players’ heights are equal to or greater than the median volleyball players’ heights

Step-by-step Explanation:

Based on the box plots given in the diagram attached below, the median height of volleyball players is 79.

Also, from the data set of basketball players, we can approximately infer that the heights from Q2, Q3, and Q4 make up 75% of the data set of the heights of basketball players.

Heights at Q2 in the data set of basketball players starts from 79, which is equal to the median height of volleyball players.

Therefore, from the displays, we can conclude that "75% of the basketball players’ heights are equal to or greater than the median volleyball players’ heights".

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