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3 years ago

Jeri drew the following model to represent the fraction. What's wrong with her model?

Mathematics

1 answer:

Shtirlitz [24]3 years ago

6 0

Answer: were is the model

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What is the volume of a cone with a radius of 2 and a height of 19?

olga55 [171]

Answer:

79.59 units ³

Step-by-step explanation:

As we know, Volume of cone = ⅓πr²h

<u>Given</u>

r = 2 units

h = 19 units

π= 22/7

$\frac{1}{3} \times \frac{22}{7} \times 2 \times 2 \times 19$

= 79.59 units ³

6 0

3 years ago

Read 2 more answers

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

The figure shown below is composed of a semicircle and a non-overlapping equilateral triangle and contains a hole that is also c

Sloan [31]

Check the picture below.

since we know the radius of the larger semicircle is 8, thus its diameter is 16, which is the length of one side of the equilateral triangle. We also know the smaller semicircle has a radius of 1/3, and thus a diameter of 2/3, namely the lenght of one side of the small equilateral triangle.

now, if we just can get the area of the larger figure and the area of the smaller one and subtract the smaller from the larger, we'll be in effect making a hole/gap in the larger and what's leftover is the shaded figure.

$\bf \stackrel{\textit{area of a semi-circle}}{A=\cfrac{1}{2}\pi r^2\qquad r=radius}~\hspace{10em}\stackrel{\textit{area of an equilateral triangle}}{A=\cfrac{s^2\sqrt{3}}{4}\qquad s=\stackrel{side's}{length}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Areas}}{\left[ \stackrel{\textit{larger figure}}{\cfrac{1}{2}\pi 8^2~~+~~\cfrac{16^2\sqrt{3}}{4}} \right]\qquad -\qquad \left[ \cfrac{1}{2}\pi \left( \cfrac{1}{3} \right)^2 +\cfrac{\left( \frac{2}{3} \right)^2\sqrt{3}}{4}\right]}$

$\bf \left[ 32\pi +64\sqrt{3} \right]\qquad -\qquad \left[ \cfrac{\pi }{18}+\cfrac{\frac{4}{9}\sqrt{3}}{4} \right] \\\\\\ \left[ 32\pi +64\sqrt{3} \right]\qquad -\qquad \left[ \cfrac{\pi }{18}+\cfrac{\sqrt{3}}{9} \right]~~\approx~~ 211.38 - 0.37~~\approx~~ 211.01$