Find the missing number in the factor tree for 20 points i hope you will help

You have 26 cards, each labeled with a letter of the alphabet. What is the probability of drawing a car that is a vowel (excludi

Alexxandr [17]

The answer is 5/26 because there are 5 vowels and 26 letters in the alphabet

6 0

3 years ago

A student draws two parabolas both parabolas cross the x axis at (-4,0) and (6,0) the y intercept of the first parabolas is (0,-

faust18 [17]

Answer:

The positive difference between the a values is 0.5

Step-by-step explanation:

we know that

Both parabolas cross the x axis at (-4,0) and (6,0)

so

The general equation is

$y=a(x+4)(x-6)$

Find the value of a in the first parabola

The y-intercept is (0,-12)

so

For $x=0, y=-12$

substitute

$y=a(x+4)(x-6)$

$-12=a(0+4)(0-6)$

$-12=a(-24)$

$a=0.5$

Find the value of a in the second parabola

The y-intercept is (0,-24)

so

For $x=0, y=-24$

substitute

$y=a(x+4)(x-6)$

$-24=a(0+4)(0-6)$

$-24=a(-24)$

$a=1$

Find the positive difference between the a values for the two functions

so

$1-0.5=0.5$

5 0

3 years ago

Read 2 more answers

Divide this decimals 5.2 divided 4

allochka39001 [22]

Answer:

1.3

Step-by-step explanation:

5.2/4 = 1.3

To check, 1.3 multiplied by four is 5.2.

7 0

3 years ago

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

If every positive number has two square roots and you can find of the side of a square window by finding a square root of the ar

butalik [34]

Every positive number has two square roots - a positive and a negative (unless the value is 0).
The length of something can only be a positive number and since there is only one positive square root, there is only one answer.

5 0

3 years ago