F(x)=1/3x+7 find inverse

We’re not done just yet! Thanks for anyone who helped, means a lot to me!

olga_2 [115]

Answer: Choice A) The number that is 5 to the left of -3 on the number line.

Explanation:

Refer to the diagram below. The value -3 is 3 units to the left of 0 on the number line. So you start at 0 and move 3 spaces to the left to arrive at -3.

Once you're at -3 on the number line, you'll move 5 more spaces to the left to arrive at -8

We can think of -3+(-5) as -3-5, both of which simplify to -8

Or we can think of it like saying -3+(-5) = -1*(3+5) = -1*8 = -8. Here I factored out a negative 1, and then added. The final result is negative since we're moving to the left in the negative territory.

6 0

3 years ago

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How do you solve this absolute value equation ? |x + 7| = 3x - 5

RUDIKE [14]

Hope this helps :) X=6

5 0

3 years ago

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5/2 times 3/2 can you help me?

rosijanka [135]

Answer:

3.75 or 3 $\frac{3}{4}$

Step-by-step explanation:

$\frac{5}{2}$ times $\frac{3}{2}$ can be multiplied as $\frac{5*3}{2*2}$ 5×3=15 and 2×2= 4 15÷4= 3.75 or 3 $\frac{3}{4}$

6 0

3 years ago

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Plz help me its due by 11:30 today plz dont report be i want a good grade in math.

marta [7]

Answer:

(3,3) for #1

Step-by-step explanation:

Graph both equations and where they intersect is your answer

3 0

3 years ago

Let X denote the length of human pregnancies from conception to birth, where X has a normal distribution with mean of 264 days a

Kaylis [27]

Answer:

Step-by-step explanation:

Hello!

X: length of human pregnancies from conception to birth.

X~N(μ;σ²)

μ= 264 day

σ= 16 day

If the variable of interest has a normal distribution, it's the sample mean, that it is also a variable on its own, has a normal distribution with parameters:

X[bar] ~N(μ;σ²/n)

When calculating a probability of a value of "X" happening it corresponds to use the standard normal: Z= (X[bar]-μ)/σ

When calculating the probability of the sample mean taking a given value, the variance is divided by the sample size. The standard normal distribution to use is Z= (X[bar]-μ)/(σ/√n)

a. You need to calculate the probability that the sample mean will be less than 260 for a random sample of 15 women.

P(X[bar]<260)= P(Z<(260-264)/(16/√15))= P(Z<-0.97)= 0.16602

b. P(X[bar]>b)= 0.05

You need to find the value of X[bar] that has above it 5% of the distribution and 95% below.

P(X[bar]≤b)= 0.95

P(Z≤(b-μ)/(σ/√n))= 0.95

The value of Z that accumulates 0.95 of probability is Z= 1.648

Now we reverse the standardization to reach the value of pregnancy length:

1.648= (b-264)/(16/√15)

1.648*(16/√15)= b-264

b= [1.648*(16/√15)]+264

b= 270.81 days

c. Now the sample taken is of 7 women and you need to calculate the probability of the sample mean of the length of pregnancy lies between 1800 and 1900 days.

Symbolically:

P(1800≤X[bar]≤1900) = P(X[bar]≤1900) - P(X[bar]≤1800)

P(Z≤(1900-264)/(16/√7)) - P(Z≤(1800-264)/(16/√7))

P(Z≤270.53) - P(Z≤253.99)= 1 - 1 = 0

d. P(X[bar]>270)= 0.1151

P(Z>(270-264)/(16/√n))= 0.1151

P(Z≤(270-264)/(16/√n))= 1 - 0.1151

P(Z≤6/(16/√n))= 0.8849

With the information of the cumulated probability you can reach the value of Z and clear the sample size needed:

P(Z≤1.200)= 0.8849

$Z= \frac{X[bar]-Mu}{Sigma/\sqrt{n} }$