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2 years ago

Please answer :) Will give brainlst)

Mathematics

1 answer:

tekilochka [14]2 years ago

4 0

C.
Some thing are on paper

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I need help with 29, 32, 35, 38, 43, 46, 49, 44, 47, 50. ASAP

gogolik [260]

29. d=-4
32. h=45
35. n=76
38. t=5
43. x=-11
46. x≈0.7
47. x≈1.6
49. d=-2
50. t=1/2

7 0

3 years ago

14 - За = -а What would be the answer?

LiRa [457]

Answer:

Step-by-step explanation:

collect like terms

14=-a+3a

2a=14

dividing both sides

a=7

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3 years ago

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Lucia earns $150 each week. She saves 12% of her earnings.

tatiyna

$18 i think............

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2 years ago

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Inputs are represented by the variable x . True or False

NNADVOKAT [17]

Answer:

true

Step-by-step explanation:

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3 years ago

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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

3 years ago