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3 years ago

PLEASE HELP QUICKLY!! Look at the graph.

Mathematics

1 answer:

Charra [1.4K]3 years ago

7 0

If the rate of change is constant, then the graph is a straight line and the function will be linear.

In this case, the rate of change is not constant so the function cannot be linear.

Answer is the last option

Her conclusion is wrong. If the rate of change is not constant, then the function cannot be linear.

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Wich one is bigger 71 ounces or 4.5 pounds

balu736 [363]

Answer:

4.5 pounds. 1 pounds is = to 16 ounces. 16 x 4.5 is 72 ounces so 4.5 pounds is 1 ounce bigger.

8 0

3 years ago

ANSWER #3 TO BE MARKED AS BRAINLIEST

12345 [234]

V=9 1/20, because when you multiply 4/5 by 4 and 1/4 by 5 to get a common denominator you get 21/20. You add the 3 and the 5, and add the one we got from the fractions and add the 1/20 to get 9 1/20

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3 years ago

Read 2 more answers

Help me please lol.......

rewona [7]

Answer:

The top one

Step-by-step explanation:

4 0

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.