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3 years ago

What is the answer for this problem for f(x)=2x+1 and g(x)=x^2-7, find (f+g)(x)

Mathematics

1 answer:

Anvisha [2.4K]3 years ago

8 0

X²-7+2x+1=(f+g)(x)
x²+2x-6
D. x² + 2x - 6

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What is 59.36 Repeated as a fraction

Bas_tet [7]

Answer:

1484/25 is 59.36 as a fraction

3 0

3 years ago

Can someone finish the whole thing ??

Romashka-Z-Leto [24]

<h3>Answer: y = (3/2)x + 0</h3>

This is the same as y = (3/2)x

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Work Shown:

Find the slope of the line through (x1,y1) = (-2,-3) and (x2,y2) = (2,3)

m = (y2 - y1)/(x2 - x1)

m = (3 - (-3))/(2 - (-2))

m = (3 + 3)/(2 + 2)

m = 6/4

m = 3/2

The slope is the fraction 3/2. This is going to be in front of the x, or to the left of the x.

----------

Plug m = 3/2 and (x1,y1) = (-2,-3) into the point slope formula. Solve for y.

y - y1 = m(x - x1)

y - (-3) = (3/2)(x - (-2))

y + 3 = (3/2)(x + 2)

y + 3 = (3/2)x + (3/2)(2)

y + 3 = (3/2)*x + 3

y + 3 - 3 = (3/2)*x + 3 - 3

y = (3/2)x + 0

The y intercept is zero. This matches up with the fact the graph crosses the y axis at y = 0.

3 0

3 years ago

Find the value of from the following right triangle.<br><br>plzzz help i will mark you as brainliest

NikAS [45]

Answer:

Step-by-step explanation:

by Pythagoras theorem,

P²+B² =H²

5²+B²=13²

B²= 13²-5²

B²= 169-25

B²= 144

B=√144

B= 12

8 0

2 years ago

The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer ra

Snezhnost [94]

Let $X$ denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by $X_1,\ldots,X_{36}$ , each independently and identically distributed with distribution $X_i\sim\mathcal N(26,7.2)$ .

You want to find

$\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)$

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to

$\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)$

Recall that if $X\sim\mathcal N(\mu,\sigma)$ , then the sampling distribution $\overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right)$ with $n$ being the size of the sample.

Transforming to the standard normal distribution, you have

$Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}$

so that in this case,

$Z=6\dfrac{\overline X-26}{7.2}$

and the probability is equivalent to

$\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)$
$=\mathbb P(Z>1.481)\approx0.0693$

5 0

2 years ago

It took Jaheim 2 hours to read the first 100 pages of a book what was his speed?

jeka94

If in 2 hours he read 200 pgs then in 1 hour will be 50 pgs

4 0

3 years ago

Read 2 more answers