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Ivahew [28]
4 years ago
6

Write the inverse function, f^-1 (x), for the function f (x) = 2x + 3

Mathematics
1 answer:
cricket20 [7]4 years ago
6 0

f(x) = 2x + 3

y = 2x + 3

Subtract the sides of the equation minus 3

y - 3 = 2x

Divided the sides of the equation by 2

\frac{y - 3}{2} = x \\

So ;

{f}^{ - 1}(x) =  \frac{x - 3}{2} \\

_________________________________

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

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Am I to assume this is for volume? The volume formula for a cylinder...oblique or not...is V = 1/3pi r^2 h.  Cavalieri's principle tells us that the formulas for both are the same. So V = 1/3*pi*100*11 and V = 366.67 pi
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It took a race car driver 3 hours and 15 minutes to go 500 miles. What was his mean rate of speed, to the nearest tenth of a mil
WINSTONCH [101]
In order to calculate rate or average speed you divide miles by time

I got 153.8 by dividing 500 by 3.25 (the .25 is because 15 minutes is 1/4 of 60) hope this helped!!
4 0
3 years ago
Find the factors<br> 4x^2+9
Vlad [161]

Step-by-step explanation:

So 4x2+9 has no linear factors with Real coefficients. It is possible to factor it with Complex coefficients.

7 0
3 years ago
X and y are normal random variables with e(x) = 2, v(x) = 5, e(y) = 6, v(y) = 8 and cov(x,y)=2. determine the following: e(3x 2y
andriy [413]

The result for the given normal random variables are as follows;

a. E(3X + 2Y) = 18

b. V(3X + 2Y) = 77

c. P(3X + 2Y < 18) = 0.5

d. P(3X + 2Y < 28) = 0.8729

<h3>What is normal random variables?</h3>

Any normally distributed random variable having mean = 0 and standard deviation = 1 is referred to as a standard normal random variable. The letter Z will always be used to represent it.

Now, according to the question;

The given normal random variables are;

E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8.

Part a.

Consider E(3X + 2Y)

\begin{aligned}E(3 X+2 Y) &=3 E(X)+2 E(Y) \\&=(3) (2)+(2)(6 )\\&=18\end{aligned}

Part b.

Consider V(3X + 2Y)

\begin{aligned}V(3 X+2 Y) &=3^{2} V(X)+2^{2} V(Y) \\&=(9)(5)+(4)(8) \\&=77\end{aligned}

Part c.

Consider P(3X + 2Y < 18)

A normal random variable is also linear combination of two independent normal random variables.

3 X+2 Y \sim N(18,77)

Thus,

P(3 X+2 Y < 18)=0.5

Part d.

Consider P(3X + 2Y < 28)

Z=\frac{(3 X+2 Y-18)}{\sqrt{77}}

\begin{aligned} P(3X + 2Y < 28)&=P\left(\frac{3 X+2 Y-18}{\sqrt{77}} < \frac{28-18}{\sqrt{77}}\right) \\&=P(Z < 1.14) \\&=0.8729\end{aligned}

Therefore, the values for the given normal random variables are found.

To know more about the normal random variables, here

brainly.com/question/23836881

#SPJ4

The correct question is-

X and Y are independent, normal random variables with E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8. Determine the following:

a. E(3X + 2Y)

b. V(3X + 2Y)

c. P(3X + 2Y < 18)

d. P(3X + 2Y < 28)

8 0
2 years ago
Find the area of the trapezoid.<br> b1=5<br> b2=7<br> h=4<br><br> PLEASE EXPLAIN TOO!
olga_2 [115]

Step-by-step explanation:

The formula for finding the area of a trapezoid is\frac{b_{1} + b_{2} }{2} ×h

Start by substituting the values given in the problem into the formula

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Now Simply/solve

\frac{12}{2} ×4

6×4

24

The area of the trapezoid is 24units^{2}

I hope this helps!!!

4 0
3 years ago
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