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2 years ago

14

Plsssssssssssssss help with 11

1 answer:

alina1380 [7]2 years ago

5 0

Answer:

1 = $3.50

( I tried my best)

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Can somebody help outtt ?? pleasee

OleMash [197]

Answer:

y=71

x=109

Step-by-step explanation:

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2 years ago

Please help me!!! Will get brainliest!!!

ratelena [41]

1. You already did it.
2. Table
3. t (years since 1990)
4. n (# of cigarettes sold)
5. (t, n)
6. You can see the distribution of the data pretty neatly. There are also many more advantages including it's easier to calculate standard deviation, easier to see the mean, mode and median, and it's also much easier to just tell the extrema of the dataset by just looking at the scattergram.

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3 years ago

Question:Find the area of he trapezoid by decomposing it into other shapes.

Ad libitum [116K]

Answer:

D) 171 cm²

Step-by-step explanation:

You don’t actually need to decompose it. You could just do it as a trapezoid:

(15+23)/2=19

19x9=171

If you want to decompose it into 2 triangles and a rectangle:

10=15x9=135

1/2(3)(9)=13.5

1/2(5)(9)=22.5

135+13.5+22.5=171 also

8 0

3 years ago

Please help. !!!!! Only if you are good at college algebra

kicyunya [14]

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3 years ago

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

<em>Probability of renting no more than 1 apartment = Probability of renting no apartment + Probability of renting 1 apartment</em>

<em />

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