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Sidana [21]
3 years ago
5

What is the value of x? Enter your answer in the box.

Mathematics
1 answer:
mariarad [96]3 years ago
4 0
In a 45-45-90 triangle, the hypotenuse = sqrt of 2 times a leg.
Therefore the legs of the 45-45-90 triangle are 6.
That is also the hypotenuse of the 30-60-90 triangle.
In a 30-60-90 triangle, the short leg is one half of the hypotenuse.
Therefore x = 3
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Ramona is counting the posts between mile markers on the highway. In 1 mile, she counts 33 posts. If the posts are evenly spaced
Naddika [18.5K]

Answer:

160 feet

Step-by-step explanation:

In order to solve this, we need to realize that the mile is being divided into 33 equal parts. Therefore our operation should look like this:

5,280 (the number of feet in 1 mile) / 33

5,280/33=160

The posts are 160 feet apart from each other.

HTH :)

5 0
2 years ago
Yesterday it rained 5/8 inch. Today it rained 1/8 inch. What is the total amount of rain for the two days?
Alecsey [184]
The total amount is 0.75 inches or 3/4 inches.
7 0
3 years ago
Montel spends $0.75 every time he plays his favorite video game. If you have $10, write and resolve an inequality to calculate h
Bumek [7]

Answer:13.333333 repeating

Step-by-step explanation:money divided by how much the games cost

3 0
3 years ago
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IceJOKER [234]

Answer:

What language is this?

3 0
2 years ago
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}.

(c) \displaystyle P(B) = \frac{9}{100}.

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

Step-by-step explanation:

<h3>(a)</h3>

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.

<h3>(b)</h3>

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}.

<h3>(c)</h3>

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}.

<h3>(d)</h3>

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
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