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3 years ago

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The table below relates the number of rats in a population to time in weeks. Use the table to write a linear equation with w as

the input variable.
P(w)=

2 answers:

miss Akunina [59]3 years ago

8 0

Answer:

$P(w)=6w+9$

Step-by-step explanation:

To find the linear equation, first we need to calculate the slope of that line, we is defined as

$m=\frac{y_{2}-y_{1} }{x_{2} -x_{1} }$

Where we need to use two points from the table: (0,9) and (4,33).

Replacing these points, we have

$m=\frac{33-9}{4-0}=\frac{24}{4} =6$

Now, we use the point-slope formula to find the equation

$y-y_{1} =m(x-x_{1} )\\y-9=6(x-0)\\y=6x+9$

Let's call $y=P(w)$ and $x=w$ .

The equation that models the given table is

$P(w)=6w+9$

salantis [7]3 years ago

7 0

Answer:

C(w) = 6w + 9

Step-by-step explanation:

Anytime 0 is given somewhere, it should be given close scrutiny. In an equation whose general form is

y = mx + b

0 will determine the y intercept immediately.

So when x = 0, y will equal

y = 0*m + 9

So b = 9

y = mx + 9 Now we need to find m

I should start using your variables.

C(w) = m*w + 9

when w = 3 then C(3) = 27

27= 3m +9

27-9 =3m + 9 -9

18 = 3m

18/3 = 3m/3

x = 6

So the complete equation is

C(w) = 6w + 9

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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

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3 years ago

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Answer:

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Step-by-step explanation:

For this problem we will use percentage calculation and make proportion:

Initial value 56 points is worth 100%

Second value 56 - 42 = 14 points of decrease is worth x%

56 : 14 = 100 : x

When we multiply external and internal members we get:

56 · x = 14 · 100 => x = 1400/56 = 25%

x = 25% percent decrease

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Step-by-step explanation:

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Step-by-step explanation:

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3 years ago