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Cerrena [4.2K]
3 years ago
6

At a concert 2/5 of the people were men, there were 3 times as many women as children. If there were 45 more men than children,

how many people were there at the concert.
Mathematics
1 answer:
Bogdan [553]3 years ago
8 0

This involves creating some equations.

First, these are what the letters I used stand for:

c = children

w = women

m = men

x = total population

Then, we must make an equation for each statement.

2/5x = m (2/5 of the people are men)

3c = w (there are 3 times as many women than children)

45+c = m (there are 45 more men than children)

Now, let's start plugging in our numbers:

x = all the men, women, and children

2/5(m+w+c) = m

2/5 (45+c +3c + c) = m     [Now simplify]

2/5 (45 + 5x) = m       [Now Distribute]

2/5(45) + (2/5)(5x) = m

2c + 18 = m    [Above we stated that there were the same amount as men as c +45]

2c + 18 = 45 +c     [Now set equal to c]

2c (-c) +18 = 45 +c (-c)

c+18 (-18) = 45 (-18)

c = 27

Now that we know that there was 27 children, we can plug 27 in for c in the other equations.

MEN = 27 + 45

WOMEN = 3(27)

CHILDREN = 27

Now add those three answers to find your total!

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Hello please help me! There are 3 questions, I will give brainliest!
Shalnov [3]

Answer:

1) There are 13 students in Jerry's study.

2) There are 39 students in Kathy's study.

3) Jerry's study is more trustworthy!

Step-by-step explanation:

1) Jerry's study is the one with the dot plot.

Now, the number of students is calculated by adding the total number of dots in the plot.

We have a total of 13 dots.

Thus, there are 13 students in Jerry's study.

2) Kathy's study is the one with the histogram.

The total number of students is gotten by adding the corresponding number of students on the y-axis for each range of distance on the x-axis.

Total number of students = 9 + 11 + 7 + 12 = 39 students

3) Jerry's study where he used a dot plot is likely to be more trustworthy because it gives exact values of the number of students for each distance represented whereas, Kathy's study where she used a histogram doesn't give exact values but just gives a range of distances for a particular number of people.

4 0
3 years ago
Plz help me on my homework
pishuonlain [190]
The answer to the problem is D
4 0
3 years ago
Read 2 more answers
1 2 3 4 5 6 7 8 9 10
daser333 [38]
<h3>Answer:</h3>

2.25

<h3>Explanation:</h3>

Consider the square ...

... (x+a)² = x² +2ax +a²

The constant term (a²) is the square of half the x-coefficient: a² = (2a/2)².

The x-coefficient in your expression is 3. The square of half that is ...

... (3/2)² = 9/4 = 2.25

Adding 2.25 to both sides gives ...

... x² +3x + 2.25 = 6 + 2.25

... (x +1.5)² = 8.25 . . . . completed square

4 0
3 years ago
How would I solve this equation?
Assoli18 [71]
2(7/2)^x = 49/2

Divide both sides by 2:

(7/2)^x = 49/4

I notice that 49/4 can be rewritten as (7/2)^2, so we now have:

(7/2)^x = (7/2)^2

The only way for this to be true is if x = 2.  Thus, we are done.

6 0
3 years ago
Suppose on a certain planet that a rare substance known as Raritanium can be found in some of the rocks. A raritanium-detector i
aleksandrvk [35]

Answer:

0.967 = 96.7% probability the rock sample actually contains raritanium

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Positive reading

Event B: Contains raritanium

Probability of a positive reading:

98% of 13%(positive when there is raritanium).

0.5% of 100-13 = 87%(false positive, positive when there is no raritanium). So

P(A) = 0.98*0.13 + 0.005*0.87 = 0.13175

Positive when there is raritanium:

98% of 13%

P(A) = 0.98*0.13 = 0.1274

What is the probability the rock sample actually contains raritanium?

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1274}{0.13175} = 0.967

0.967 = 96.7% probability the rock sample actually contains raritanium

7 0
2 years ago
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