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3 years ago

Find the area. Of this shape. Please help fast!

Mathematics

2 answers:

topjm [15]3 years ago

8 0

Answer:

152in^2

Step-by-step explanation:

12x6=72

8x10=80

80+72=152

skelet666 [1.2K]3 years ago

8 0

Answer:

152 in

Step-by-step explanation:

Use the attachment to see what I mean by shape A and B:

Shape A's area is : 6×20 = 120 in

Shape B's area is : 8×4=32 in

Adding the two shapes gives you a total of 152 in

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Andrews [41]

Books left=2000-200=1800

ATQ

$\\ \sf\longmapsto 1800x+10=0.08x$

$\\ \sf\longmapsto 1800x-0.08x=-10$

$\\ \sf\longmapsto x(1800-0.08)=-10$

$\\ \sf\longmapsto x(1799.92)=-10$

$\\ \sf\longmapsto x=|\dfrac{10}{1799.92}|$

$\\ \sf\longmapsto x=0.005$

Now

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2 years ago

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Step-by-step explanation:

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3 years ago

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masya89 [10]

Answer:

$P(X\leq 1) = 0.331$

Step-by-step explanation:

Given

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Required

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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

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