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3 years ago

Given f(x)=3x-5, solve for x when f(x)=1

Mathematics

2 answers:

alekssr [168]3 years ago

3 0

Answer:

x = -2

Step-by-step explanation:

f(x) = 3x - 5

= 3(1) - 5

= 3 - 5

x = -2

solniwko [45]3 years ago

3 0

Answer:

-2

Step-by-step explanation:

substitute value of x in the equation.

f(x) = 1 = f(1)

= 3(1) - 5

= 3 - 5

= -2

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3. Consider the following information.P(Z) = 0.73P(A) = 0.25P(Z and A) = 0.15Part A: Determine P(Z|A).Part B: Determine P(AZ).I

krok68 [10]

Ok, so:

For Part A, we have: P(Z|A)=P(Z and A)/P(A)

And if we replace, we got:

P(Z|A) = (0.15)/(0.25) and this is equal to 0.6.

For Part B, we have: P(A|Z)=P(Z and A)/P(Z)

P(A|Z) = (0.15)/(0.73) and this is equal to 0.205.

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2 years ago

The prism is completely filled with 1750 cubes that have edge length of 1/5 feet. What is the volume of the prism,

Taya2010 [7]

We know that
the volume one cube=(1/5)*(1/5)*(1/5)=(1/125) ft³
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the answer is 14 ft³

7 0

3 years ago

The average birth weight of elephants is 200 pounds. Assume that the distribution of birth weights is Normal with a standard dev

zhenek [66]

The birth weight of elephants at the 85th percentile is 262.16 pounds

<h3>What is the standard deviation?</h3>

Standard deviation is defined as the amount of variation or the deviation of the numbers from each other.

Given that:-

The average birth weight of elephants is 200 pounds $\mu$ = 200

The deviation of 60 pounds. $\sigma$ = 60

The birth weight of elephants at the 85th percentile. z = 85%= 1.036

Now from the formula of z-score:-

$z=\dfrac{x-\mu}{\sigma}\\\\\\z=\dfrac{x-200}{60}\\\\\\1.036\times 60=x-200\\\\\\x=\ \ 262.16\ pounds$

Hence the birth weight of elephants at the 85th percentile is 262.16 pounds

To know more about standard deviation follow

brainly.com/question/475676

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5 0

2 years ago

Let X have the uniform distribution U(0, 2) and let the conditional distribution of Y , given that X = x, be U(0, x). Find the j

Elina [12.6K]

Answer:

$f(x,y) = \frac{1}{x} \frac{1}{2}= \frac{1}{2x} , 0\leq x \leq 2 , 0\leq y \leq x$

$E(Y|x) = \int_{x=y}^2 y \frac{1}{x} dx= y ln x \Big|_{x=y}^2 =y ln 2 -y ln y = y(1-lny) \$

Step-by-step explanation:

We have two random variables X and Y. $X \sim Unif(0,2)$ and given that X=x, Y has uniform distribution (0,x)

From the definition of the uniform distribution we have the densities for each random variable given by:

$f_X (x) =\frac{1}{2} , 0\leq x\leq 2$

$f_{Y|X} (y|x) = \frac{1}{x}, 0\leq y \leq x$

And on this case we can find the joint density with the following formula:

$f(x,y) = f_{Y|X}(y|x) f_X (x)$

And multiplying the densities we got this:

$f(x,y) = \frac{1}{x} \frac{1}{2}= \frac{1}{2x} , 0\leq x \leq 2 , 0\leq y \leq x$

Now with the joint density we can find the expected value E(Y|x) with the following formula:

$E(Y|x) = \int y f_{Y|X}(y|x)dx$

And replacing we got:

$E(Y|x) = \int_{x=y}^2 y \frac{1}{x} dx= y ln x \Big|_{x=y}^2 =y ln 2 -y ln y = y(1-lny) \$

5 0

4 years ago

Pls help!<br><br> 3x-4+71+80=180

juin [17]

Answer:

the answer is x=11

3 0

3 years ago