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2 years ago

YA'LL PLEASE HELP I REALLYYY NEED HELP!!!!! (all links for answers will be deleted)

Mathematics

1 answer:

Vinil7 [7]2 years ago

8 0

Answer:

(x, y) → (x – 8, y)

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2 years ago

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IgorC [24]

Answer:

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Step-by-step explanation:

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2 years ago

Find the volume of this sphere. Use 3 for TT. V [?]in3 V = Tr3 8 in

frutty [35]

Answer:

The volume of the sphere is 2048 in³.

Step-by-step explanation:

The volume of a sphere is given by:

$V = \frac{4}{3}\pi r^{3}$

Where:

r: is the radius = 8 in

Having the radius and by using 3 for π, the volume is:

$V = \frac{4}{3}*3 (8 in)^{3} = 2048 in^{3}$

Therefore, the volume of the sphere is 2048 in³.

I hope it helps you!

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2 years ago

If f(x) = e^[3ln(x^2)], then f'(x) = ?

kherson [118]

You might need to give out the choices, if there are any. thanks for asking.

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2 years ago

Read 2 more answers

An apartment complex rents an average of 2.3 new units per week. If the number of apartment rented each week Poisson distributed

masya89 [10]

Answer:

$P(X\leq 1) = 0.331$

Step-by-step explanation:

Given

Poisson Distribution;

Average rent in a week = 2.3

Required

Determine the probability of renting no more than 1 apartment

A Poisson distribution is given as;

$P(X = x) = \frac{y^xe^{-y}}{x!}$

Where y represents λ (average)

y = 2.3

Probability of renting no more than 1 apartment = Probability of renting no apartment + Probability of renting 1 apartment

Using probability notations;

$P(X\leq 1) = P(X=0) + P(X =1)$

Solving for P(X = 0) [substitute 0 for x and 2.3 for y]

$P(X = 0) = \frac{2.3^0 * e^{-2.3}}{0!}$

$P(X = 0) = \frac{1 * e^{-2.3}}{1}$

$P(X = 0) = e^{-2.3}$

$P(X = 0) = 0.10025884372$

Solving for P(X = 1) [substitute 1 for x and 2.3 for y]

$P(X = 1) = \frac{2.3^1 * e^{-2.3}}{1!}$

$P(X = 1) = \frac{2.3 * e^{-2.3}}{1}$

$P(X = 1) =2.3 * e^{-2.3}$

$P(X = 1) = 2.3 * 0.10025884372$

$P(X = 1) = 0.23059534055$

$P(X\leq 1) = P(X=0) + P(X =1)$

$P(X\leq 1) = 0.10025884372 + 0.23059534055$

$P(X\leq 1) = 0.33085418427$

$P(X\leq 1) = 0.331$

Hence, the required probability is 0.331

6 0

2 years ago