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3 years ago

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A total of 132 people attended the Lane Family Reunion. The total cost of the reunion was $5,544. Everyone contributed the same

amount. How much was each person's contribution?

2 answers:

Andrei [34K]3 years ago

5 0

Answer:

Each person contributed 42 dollars.

Step-by-step explanation:

divide 5544 by 132

umka2103 [35]3 years ago

4 0

<h2>Answer</h2><h3>42</h3>

<h2>step by step explanation </h2>

divide : 5544÷132
Answer : 42

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Side AB of square ABCD is parallel to the Y axis and the perimeter of ABCD is 36. If the coordinates of point B is (5,17) . What

Zarrin [17]

<h3>Answer: (C) (14,8)</h3>

============================================

Explanation:

The perimeter of the square is 36, so each side length is 36/4 = 9 units.

Point B is located at (5,17). We move down 9 units to get to (5,8), which is the location of point A. Then we move 9 units to the right to arrive at (14,8) which is point D's location.

Or we could go from B = (5,17) to C = (14,17) and then to D = (14,8). Each time we move 9 units.

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3 years ago

Figure EFGH on the grid below represents a trapezoidal plate at its starting position on a rotating platform:

Murrr4er [49]

Answer: A is correct

Step-by-step explanation: I took the test and I got 100

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3 years ago

If the numbers 4, 5 and 6 are each used exactly once to replace the letters in the expression A ( B − C ), what is the least pos

guapka [62]

Answer:

The least possible result is <em>-10</em>.

Step-by-step explanation:

Given the numbers 4, 5 and 6 are to be chosen one of the letters A, B or C.

First of all,

Let A = 4, B = 5 and C = 6

$A(B-C) = 4 \times (5-6) = 4 \times -1 = -4$

Let A = 4, B = 6 and C = 5

$A(B-C) = 4 \times (6-5) = 4 \times 1 = 4$

Let A = 5, B =4 and C = 6

$A(B-C) = 5 \times (4-6) = 5 \times -2 = -10$

Let A = 5, B = 6 and C = 4

$A(B-C) = 4 \times (6-4) = 4 \times 2 = 8$

Let A = 6, B = 4 and C = 5

$A(B-C) = 6 \times (4-5) = 6 \times -1 = -6$

Let A = 6, B = 5 and C = 4

$A(B-C) = 6 \times (6-5) = 6 \times 1 = 6$

Summarizing the above values in the form of a table:

$\begin{center}\begin{tabular}{ c c c c}A & B & C & A(B-C)\\ 4 & 5 & 6 & -4\\ 4 & 6 & 5 & 4\\ 5 & 4 & 6 & -10\\ 5 & 6 & 4 & 10\\ 6 & 4 & 5 & -6\\ 6 & 5 & 4 & 6\end{tabular}\end{center}$

So, the least possible result is <em>-10</em>.

3 0

3 years ago

Social Sciences Alcohol Abstinence The Harvard School of Public Health completed a study on alcohol consumption on college campu

Scilla [17]

Answer:

a) There is a 6.69% probability that a randomly selected female student abstains from alcohol.

b) If a randomly selected female student abstains from alcohol, there is a 82.87% probability that she attends a coeducational college.

Step-by-step explanation:

This is a probability problem:

We have these following probabilities:

-20.7% of a woman attending an all-women college abstaining from alcohol.

-6% of a woman attending a coeducational college abstaining from alcohol.

-4.7% of a woman attending an all-women college

- 100%-4.7% = 95.3% of a woman attending a coeducational college.

(a) What is the probability that a randomly selected female student abstains from alcohol?

$P = P_{1} + P_{2}$

$P_{1}$ is the probability of a woman attending an all-women college being chosen and abstaining from alcohol. There is a 0.047 probability of a woman attending an all-women college being chosen and a 0.207 probability that she abstain from alcohol. So:

$P_{1} = 0.047*0.207 = 0.009729$

$P_{2}$ is the probability of a woman attending a coeducational college being chosen and abstaining from alcohol. There is a 0.953 probability of a woman attending a coeducational college being chosen and a 0.06 probability that she abstain from alcohol. So:

$P_{2} = 0.953*0.06 = 0.05718$

So, the probability of a randomly selected female student abstaining from alcohol is:

$P = P_{1} + P_{2} = 0.009729 + 0.05718 = 0.0669$

There is a 6.69% probability that a randomly selected female student abstains from alcohol.

(b) If a randomly selected female student abstains from alcohol, what is the probability she attends a coedücational colege?

<em>This can be formulated as the following problem:</em>

<em>What is the probability of B happening, knowing that A has happened.</em>

Here:

<em>What is the probability of a woman attending a coeducational college, knowing that she abstains from alcohol.</em>

It can be calculated by the following formula:

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

We have the following probabilities:

$P(B)$ is the probability of a woman from a coeducational college being chosen. So $P(B) = 0.953$

$P(A/B)$ is the probability of a woman abstaining from alcohol, given that she attends a coeducational college. So $P(A/B) = 0.06$

$P(A)$ is the probability of a woman abstaining from alcohol. From a), $P(A) = 0.0669$

So, the probability that a randomly selected female student attends a coeducational college, given that she abstains from alcohol is:

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{(0.953)*(0.06)}{(0.0669)} = 0.8287$

If a randomly selected female student abstains from alcohol, there is a 82.87% probability that she attends a coeducational college.

4 0

3 years ago

charle [14.2K]

Here mine. I did this for my AP art

6 0

3 years ago