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2 years ago

How to write 9,009,009 in word form

Mathematics

2 answers:

svetoff [14.1K]2 years ago

8 0

Nine-million-nine-thousand-and nine Hope this is useful/helpful for you !! ^_^ !!

DaniilM [7]2 years ago

7 0

Nine million nine thousand and nine

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At the start of a race, a runner’s velocity changes from 0 to 4.0 m/s. It takes him 2 seconds to speed up. His acceleration is _

maksim [4K]

To find the acceleration we divide the change of velocity by the change of time.
Change of velocity: 4 m/s
Change of time: 2 s

Acceleration = $\frac{4}{2} = 2$

So the answer: his acceleration is 2 m/s^2

8 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

2 years ago

.<br> 5. How ounces are in a gallon of milk?

Natalka [10]

Step-by-step explanation:

it would be 128 ounces

hope i helped

3 0

2 years ago

What is the degree of 5x3 + 2x2y3z?

Ulleksa [173]

Answer:

stylo fabric customer care 9903160322##8944928267

7 0

3 years ago

Read 2 more answers

23rd term: -4,-7,-10,-13

Mashutka [201]

5 : -16
6 : -19
7 : -22
8 : -25
9 : -28
10: -31
11 : -34
12: -37
13: -40
14 : -43
15 : -46
16 : - 49
17 : - 52
18 : - 55
19 : - 58
20: - 61
21 : - 64
22: -67
23: -70

23rd term : is -70

3 0

3 years ago