Find the product. list the units

Can someone help me with this

arsen [322]

Answer:

1. 40 members

2. 4 cheerleaders

3. 40 marbles

4. 2 marbles

5. 8

6. 24

Step-by-step explanation:

1. cross multiply 12/x=30/100

2. cross multiply 12/x=10/100

3. 18/x=45/100

4. x/40=5/100

5. we know 5% is 2 marbles. multiply by 4 to get 8 marbles.

6. 20%=8 marbles. 60/20= 3. 3 x 8= 24 marbles.

6 0

2 years ago

51.03 is 88% of what number

Tcecarenko [31]

I would use a proportion

51.03/88% = x/100%

Which is saying: if 51.03 is 88% of the number, what is 100% of the number

Cross multiply to get x=57.99

5 0

2 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.