What is lower 910/7 or 91/3/5

If h(x) = x – 7 and g(x) = x2, which expression is equivalent to (5 – 7)2 (5)2 – 7 (5)2(5 – 7) (5 – 7)x2

rodikova [14]

E are given with two functions: h(x) = x – 7 and g(x) = x2. In this problem, we are asked to determine the value of g(h(5)). first we get g(h(x)) which is equal to (x-7)2, when x is substituted by 5, then g(h(5)) = (5-7)^2 equal to 4

3 0

3 years ago

PLS HELP ME ON THIS QUESTION I WILL MARK YOU AS BRAINLIEST IF YOU KNOW THE ANSWER!!

devlian [24]

Answer:

C. 98

Step-by-step explanation:

If you add all of his previous test scores and go through the list of numbers you get C.98 as the answer. I added all of his test scores plus 98 which equals 630 divided by 7 (because if he got a 98 there would be 7 test scores) you would get 90 or an A average.

3 0

3 years ago

Read 2 more answers

Help: see picture for information. How much will Jerry have to pay if he wants to pay as little money as possible?

wolverine [178]

105 is the correct answer

5 0

4 years ago

Read 2 more answers

Can someone explain to me how to do this?

elena55 [62]

In the slope intercept form equation of a line (y=ax+b) a represents the slope, and b represents the intercept, which is -5. Thus, we know y=ax-5. The slope can be found by creating a fraction with the change in y value on top, and the change of x value on the bottom. If we have two points (4, 1) and (0, -5) we can put this into a fraction - $\frac{-5-1}{0-4}$ , which simplifies down to -6/-4, or 3/2. Thus the answer is f(x)= $\frac{3}{2}$ x-5<span />

6 0

4 years ago

A lamp has two bulbs, each of a type with average lifetime 1400 hours. Assuming that we can model the probability of failure of

liq [111]

Answer:

The probability of failure of both the bulbs is 0.4323.

Step-by-step explanation:

For an exponential distribution the distribution is given by

$f(x,\lambda )=\int_{0}^{x }\lambda e^{-\lambda x}dx$

The value of λ is related to the mean μ as λ=1/μ,

Let us denote the 2 bulbs by X and Y thus the probability distribution of the 2 bulbs is as under

$P(X)=\int_{0}^{x }\lambda _{X}e^{-\lambda _{X}x}dx$

Similarly for the bulb Y the distribution function is given by

$P(Y)=\int_{0}^{y }\lambda _{Y}e^{-\lambda _{Y}y}dy$

Thus the probability for both the bulbs to fail within 1500 hours is

$P(E)=\int_{0}^{1500}\int_{0}^{1500}\frac{1}{1400}e^{\frac{-x}{1400}}\cdot \frac{1}{1400}e^{\frac{-y}{1400}}dxdy\\\\P(E)=\frac{1}{1400^2}(\int_{0}^{1500}\int_{0}^{1500}e^{\frac{-x}{1400}}\cdot e^{\frac{-y}{1400}}dxdy)\\\\P(E)=\frac{1}{1400^2}(\int_{0}^{1500}e^{\frac{-x}{1400}}dx)\cdot (\int_{0}^{1500}e^{\frac{-y}{1400}}dy)\\\\P(E)=\frac{1}{1400^{2}}\times 920.473\times 920.473\\\\\therefore P(E)=0.4323$

6 0

3 years ago