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3 years ago

If I have a Pentagon shape and it's 33 centimeters long, what would be my Perimeter?

Mathematics

1 answer:

Oksana_A [137]3 years ago

4 0

The answer is 165. If you add 33 five times you get 165. Or you could do what i did and do 33 times 5.

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Points!!!

sattari [20]

(1/40) - (1/x) = (1/60)
x = 1/(1/40) - (1/60)) =
<span>A.) 120</span> minutes
Its similar to the other question.

7 0

2 years ago

Which of the following is an even function? A.f(x) =(x-1)^2 b. F(x)=8x c.f(x) =x^2-x d.f(x)=7

pentagon [3]

Answer:

Step-by-step explanation:

A.f(x) =(x-1)^2 is even because of that even exponent, 2.

b. F(x)=8x is odd because x has the exponent 1.

c.f(x) =x^2-x is neither even nor odd

d.f(x)=7 is even because the exponent is even: 7^0

4 0

3 years ago

Lionel’s work crew packages 150,087 bottles of water every day. If the crew works 5 days, how many bottles can they package in a

yawa3891 [41]

Answer:

im sorry this might not be exactly correct but the answer can be 1,050,609

Step-by-step explanation:

1 week= 7 days so

150,087 × 7 = 1,050,609

‍♀️

3 0

2 years ago

During which two months do both the Northern and Southern Hemispheres receive the same amount of energy from the Sun?

Drupady [299]

Answer: The correct answer is March and September

pls mark brainliest

6 0

2 years ago

Read 2 more answers

The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(

Umnica [9.8K]

Answer:

(a) $A_1$ and $A_2$ are indeed mutually-exclusive.

(b) $\displaystyle P(A_1\; \cap \; B) = \frac{1}{20}$ , whereas $\displaystyle P(A_2\; \cap \; B) = \frac{1}{25}$ .

(d) $\displaystyle P(A_1 \; |\; B) \approx \frac{5}{9}$ , whereas $P(A_1 \; |\; B) = \displaystyle \frac{4}{9}$

Step-by-step explanation:

$P(A_1 \; \cap \; A_2) = 0$ means that it is impossible for events $A_1$ and $A_2$ to happen at the same time. Therefore, event $A_1$ and $A_2$ are mutually-exclusive.

By the definition of conditional probability:

$\displaystyle P(B \; | \; A_1) = \frac{P(B \; \cap \; A_1)}{P(B)} = \frac{P(A_1 \; \cap \; B)}{P(B)}$ .

Rearrange to obtain:

$\displaystyle P(A_1 \; \cap \; B) = P(B \; |\; A_1) \cdot P(A_1) = 0.25 \times 0.20 = \frac{1}{20}$ .

Similarly:

$\displaystyle P(A_2 \; \cap \; B) = P(B \; |\; A_2) \cdot P(A_2) = 0.80 \times 0.05 = \frac{1}{25}$ .

Note that:

$\begin{aligned}P(A_1 \; \cup \; A_2) &= P(A_1) + P(A_2) - P(A_1 \; \cap \; A_2) = 0.20 + 0.80 = 1\end{aligned}$ .

In other words, $A_1$ and $A_2$ are collectively-exhaustive. Since $A_1$ and $A_2$ are collectively-exhaustive and mutually-exclusive at the same time:

$\displaystyle P(B) = P(B \; \cap \; A_1) + P(B \; \cap \; A_2) = \frac{1}{20} + \frac{1}{25} = \frac{9}{100}$ .

By Bayes' Theorem:

$\begin{aligned} P(A_1 \; |\; B) &= \frac{P(B \; | \; A_1) \cdot P(A_1)}{P(B)} \\ &= \frac{0.25 \times 0.20}{9/100} = \frac{0.05 \times 100}{9} = \frac{5}{9}\end{aligned}$ .

Similarly:

$\begin{aligned} P(A_2 \; |\; B) &= \frac{P(B \; | \; A_2) \cdot P(A_2)}{P(B)} \\ &= \frac{0.05 \times 0.80}{9/100} = \frac{0.04 \times 100}{9} = \frac{4}{9}\end{aligned}$ .

6 0

2 years ago