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3 years ago

The height of a box is 7 the length is three inches more than the width find the volume is 126

Mathematics

1 answer:

babunello [35]3 years ago

6 0

Answer:

<h2>answer c: 10</h2>

Step-by-step explanation:

v=H*W*L

280=7*W*(6+W)

280=42W+7W^2

40=6w+w^2

W^2+6W-40=0

(W-4)(W+10)=0

W=4

We can only use positive results

L=6+W

L=4+6=10

=10

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A selective college would like to have an entering class of 950 students. Because not all students who are offered admission acc

pogonyaev

Answer:

a) The mean is 900 and the standard deviation is 15.

b) 100% probability that at least 800 students accept.

c) 0.05% probability that more than 950 will accept.

d) 94.84% probability that more than 950 will accept

Step-by-step explanation:

We use the normal approximation to the binomial to solve this question.

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 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.

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

(a) What are the mean and the standard deviation of the number X of students who accept?

$n = 1200, p = 0.75$ . So

$E(X) = np = 1200*0.75 = 900$

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1200*0.75*0.25} = 15$

The mean is 900 and the standard deviation is 15.

(b) Use the Normal approximation to find the probability that at least 800 students accept.

Using continuity corrections, this is $P(X \geq 800 - 0.5) = P(X \geq 799.5)$ , which is 1 subtracted by the pvalue of Z when X = 799.5. So

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

$Z = \frac{799.5 - 900}{15}$

$Z = -6.7$

$Z = -6.7$ has a pvalue of 0.

1 - 0 = 1

100% probability that at least 800 students accept.

(c) The college does not want more than 950 students. What is the probability that more than 950 will accept?

Using continuity corrections, this is $P(X \geq 950 - 0.5) = P(X \geq 949.5)$ , which is 1 subtracted by the pvalue of Z when X = 949.5. So

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

$Z = \frac{949.5 - 900}{15}$

$Z = 3.3$

$Z = 3.3$ has a pvalue of 0.9995

1 - 0.9995 = 0.0005

0.05% probability that more than 950 will accept.

(d) If the college decides to increase the number of admission offers to 1300, what is the probability that more than 950 will accept?

Now n = 1300. So

$E(X) = np = 1300*0.75 = 975$

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1200*0.75*0.25} = 15.6$

Same logic as c.

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

$Z = \frac{949.5 - 975}{15.6}$

$Z = -1.63$

$Z = -1.63$ has a pvalue of 0.0516

1 - 0.0516 = 0.9484

94.84% probability that more than 950 will accept

5 0

3 years ago

35 points!! How much does a fish weigh if its tail weighs 4 kilograms, it's head weighs as much as it's tail and half its body,

NeX [460]

Let t, h, b represent the weighs of tail, head, and body, respectively.

t = 4 . . . . given

h = t + b/2 . . . . the head weighs as much as the tail and half the body

b/2 = h + t . . . . half the body weighs as much as the head and tail

_____

Substituting for b/2 in the second equation using the expression in the third equation, we have

... h = t + (h + t)

Subtracting h from both sides gives

... 0 = 2t . . . . . . in contradiction to the initial statement about tail weight.

Conclusion: there's no solution to the problem given here.

4 0

3 years ago

What is the common difference between successive terms in the sequence?

anzhelika [568]

Look at each term. How does the next one change? what is added?

The answer is 6, because there is an increase of 6 for each term

7 0

3 years ago

Read 2 more answers

An irrational number is one that simply can not be worked into any equation.

Fudgin [204]

Answer:

it can be use in equation

An irrational number is a number that cannot be expressed as a fraction for any integers and. . Irrational numbers have decimal expansions that neither terminate nor become periodic.

5 0

2 years ago

Please help me on 7&8 and please show your work :)

Llana [10]

7. 6/12 or 1/2 | 8. 5/12

5 0

3 years ago