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Hitman42 [59]
3 years ago
15

Help please this is hard

Mathematics
1 answer:
Yuri [45]3 years ago
3 0
The answer is 0,4,known ás the answer C
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36/48=6/8=3/4 so 75% completed training
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A rectangular countertop has a perimeter of 26 feet. its area is 30 square feet. what are the dimensions of the counter?
Sladkaya [172]
P=26=2*(l+L)
⇒ l+L=26:2=13 feet
l=3, L=10

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4 0
3 years ago
A class of 40 students elected a class president. There were 12 votes for Candidate A, and 18 votes for Candidate B.
Helen [10]
Hey there, Bulgogi!

The correct answer will be the first one since [I'm assuming] you know that to calculate the percentage of something, it's the amount of that something divided by the total amount but since only 30 students voted [12+18] multiplied by 100.

So, we are calculating the percentage on votes and 12 students voted for Candidate A so our Numerator will be 12 and our Denominator will be the total amount of student who voted, so 30 [or 12+18]

Once put into a fraction, we will get \frac{12}{12+18} which would be equals to 2/5 or 0.4. Now we multiply by 100 and we get a total of 40% for Candidate A.

Thank you for using Brainly.
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4 0
4 years ago
Calculate the value of 1/3 a = -5
sladkih [1.3K]

Answer:

- 15

Step-by-step explanation:

\frac{1}{3} a =  - 5 \\  \\ a =  - 5 \times 3 \\  \\ a =  - 15

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}.

(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
3 years ago
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