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3 years ago

What are the equivalent fractions with a common denominator for 2/5 and 1/2?

Mathematics

2 answers:

Svetlanka [38]3 years ago

4 0

So 1/2 is 5/10 and 2/5 is 4/10 common denonominator is 10 ( i think )

kirill115 [55]3 years ago

3 0

4/10 and 5/10 with 10 being the common <span>denominator :)</span>

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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

Can someone please clearly and correctly solve and show me how to work out QUESTION 3 AND 4??

vredina [299]

There are 2 options to solve that.
1. The first one is by derivatives.
f(x)=x^2+12x+36
f'(x)=2x+12
then you solve that for f'(x)=0
0=2x+12
x=(-6)
you have x so for (-6) solve the first equation, then you find y
y=(-6)^2+12*(-6)+36=(-72)
so the vertex is (-6, -72)
2. The second option is to solve that by equations:
for x we have:
x=(-b)/2a
for that task we have
b=12
a=1
x=(-12)/2=(-6)
you have x so put x into the main equation
y=(-6)^2+12*(-6)+36=(-72)
and we have the same solution: vertex is (-6, -72)

For next task, I will use the second option:
y=x^2-6x
x=(-b)/2a
for that task we have
b=(-6)
a=1
x=(6)/2=3
you have x so put x into the main equation
y=3^2+(-6)*3=(--9)
and we have the same solution: vertex is (3, -9)

3 0

3 years ago

Write an equation of a line with the given slope and y-intercept. m = –5, b = –3. A. y = –5x – 3 . B. y = –5x + 3 . C. y = 5x –

True [87]

Y = mx + b
y = -5x -3

5 0

3 years ago

Read 2 more answers

Prove that:<br><br>cos20°cos40°cos80°=1/8

Naya [18.7K]

Answer:

see explanation