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4 years ago

Shade more than 11/100 and less than 1/4. Write the value of the shaded part as decimal and a fraction

Mathematics

1 answer:

devlian [24]4 years ago

3 0

20/100 and 0.20 because 20 is in middle of 11 and25

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3 years ago

Suppose you are given that a=b+2 and b+2=5. What can you prove by using these statements and the Transitive Property?

Nataliya [291]

You can prove that a=5, b=3, 5=3+2 and a=b+2.

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3 years ago

The University of Washington claims that it graduates 85% of its basketball players. An NCAA investigation about the graduation

Nonamiya [84]

Probabilities are used to determine the chances of events

The given parameters are:

Sample size: n = 20
Proportion: p = 85%

<h3>(a) What is the probability that 11 out of the 20 would graduate? </h3>

Using the binomial probability formula, we have:

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

So, the equation becomes

$P(x = 11) = ^{20}C_{11} \times (85\%)^{11} \times (1 - 85\%)^{20 -11}$

This gives

$P(x = 11) = 167960 \times (0.85)^{11} \times 0.15^{9}$

$P(x = 11) = 0.0011$

Express as percentage

$P(x = 11) = 0.11\%$

Hence, the probability that 11 out of the 20 would graduate is 0.11%

<h3>(b) To what extent do you think the university’s claim is true?</h3>

The probability 0.11% is less than 50%.

Hence, the extent that the university’s claim is true is very low

<h3>(c) What is the probability that all 20 would graduate? </h3>

Using the binomial probability formula, we have:

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

So, the equation becomes

$P(x = 20) = ^{20}C_{20} \times (85\%)^{20} \times (1 - 85\%)^{20 -20}$

This gives

$P(x = 20) = 1 \times (0.85)^{20} \times (0.15\%)^0$

$P(x = 20) = 0.0388$

Express as percentage

$P(x = 20) = 3.88\%$

Hence, the probability that all 20 would graduate is 3.88%

<h3>(d) The mean and the standard deviation</h3>

The mean is calculated as:

$\mu = np$

So, we have:

$\mu = 20 \times 85\%$

$\mu = 17$

The standard deviation is calculated as:

$\sigma = np(1 - p)$

So, we have:

$\sigma = 20 \times 85\% \times (1 - 85\%)$

$\sigma = 20 \times 0.85 \times 0.15$

$\sigma = 2.55$

Hence, the mean and the standard deviation are 17 and 2.55, respectively.