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4 years ago

Write each percent as a fraction in simplest form

Mathematics

1 answer:

sattari [20]4 years ago

3 0

Im sorry, I cant see what it is you want me to do??

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There are 1493 men and 1438 women in the audience at a concert. What is the percentage of the audience that are men

Advocard [28]

To calculate this answer you have to sum up the amount of women and amount of men at the concert: 1493 + 1438 = 2931.

Now you have the total amount of people. To get the precentage of the audience that are men, you do this: 1493/2931. (you divide the amount of men with the amount of every person present).

1493/2931 = 0.5093

you multiply 0.5093 by 100

and you get 50.93%. And that is the final answer.

8 0

3 years ago

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Y^2 -40y +c find the value of c that completes the square

o-na [289]

Answer:

$c = 400$

Step-by-step explanation:

${y}^{2} - 40y + 400 = {(y - 20)}^{2}$

5 0

3 years ago

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Use a timeline how much time elapses from 9:45 a.m. to 11:10 a.m.

Neporo4naja [7]

Answer:

1 hour 25 minutes

(or)

85 minutes

Step-by-step explanation:

4 0

3 years ago

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1500 customers hold a VISA card; 500 hold an American Express card; and, 75 hold a VISA and an American Express. What is the pro

alex41 [277]

Answer:

There is 15% probability that a customer chosen at random holds a VISA card, given that the customer has an American Express card.

$P(VISA \:| \:AE) = 15\%\\$

Step-by-step explanation:

Number of customers having a Visa card = 1,500

Number of customers having an American Express card = 500

Number of customers having Visa and American Express card = 75

Total number of customers = 1,500 + 500 = 2,000

We are asked to find the probability that a customer chosen at random holds a VISA card, given that the customer has an American Express card.

This problem is related to conditional probability which is given by

$P(A \:| \:B) = \frac{P(A \:and \:B)}{P(B)}$

For the given problem it becomes

$P(VISA \:| \:AE) = \frac{P(VISA \:and \:AE)}{P(AE)}$

The probability P(VISA and AE) is given by

P(VISA and AE) = 75/2000

P(VISA and AE) = 0.0375

The probability P(AE) is given by

P(AE) = 500/2000

P(AE) = 0.25

Finally,

$P(VISA \:| \:AE) = \frac{P(VISA \:and \:AE)}{P(AE)}\\\\P(VISA \:| \:AE) = \frac{0.0375}{0.25}\\\\P(VISA \:| \:AE) = 0.15\\\\P(VISA \:| \:AE) = 15\%\\$

Therefore, there is 15% probability that a customer chosen at random holds a VISA card, given that the customer has an American Express card.

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3 years ago

How many solutions does x^2-6x=-14 have

valina [46]

Answer:

it has 0 solution

Step-by-step explanation:

b^2-4ac<0=0 real # solution

brainliest please

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3 years ago

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