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2 years ago

Write a real world problem for the equation 11x =385.

1 answer:

7 0

Real world problem for 11x = 385

Sam brought home 385 packs of string for the party. At the party, Sam invites 11 people, and wants to give each person the same amount of string. How much string does each person get?

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EASY QUESTION.

VMariaS [17]

Biomass is plant or animal material used for energy production (electricity or heat), or in various industrial processes as raw substance for a range of products

3 0

3 years ago

I need help with this question

Ghella [55]

Answer:

$18.63

Step-by-step explanation:

15% of 45 is equal to 45 times .15, which is equal to 6.75.

45+6.75 = 51.75

Ben resold the guitar for $51.75

36% of 51.75 is equal to 51.75*.36 or 18.63

Ben spent $18.63 on books

8 0

2 years ago

Read 2 more answers

The manager suggests that Darryl should purchase the deli’s bread from Store D because they sell the most loaves of bread for th

stealth61 [152]

Answer:

yes

Step-by-step explanation:

because it sells the best price

5 0

3 years ago

Read 2 more answers

A: {71,73,79,83,87} B:{57,59,61,67}

Jobisdone [24]

Answer:

$\frac{3}{5}$ .

Step-by-step explanation:

We have been given two sets as A: {71,73,79,83,87} B:{57,59,61,67}. We are asked to find the probability that both numbers are prime, if one number is selected at random from set A, and one number is selected at random from set B.

We can see that in set A, there is only one non-prime number that is 87 as it is divisible by 3.

So there are 4 prime number in set A and total numbers are 5.

$P(\text{Prime number from A})=\frac{4}{5}$

We can see that in set B, there is only one non-prime number that is 57 as it is divisible by 3.

So there are 3 prime number in set B and total numbers are 4.

$P(\text{Prime number from B})=\frac{3}{4}$

Now, we will multiply both probabilities to find the probability that both numbers are prime. We are multiplying probabilities because both events are independent.

$P(\text{Prime number from A and B})=\frac{4}{5}\times \frac{3}{4}$

$P(\text{Prime number from A and B})=\frac{1}{5}\times \frac{3}{1}$

$P(\text{Prime number from A and B})=\frac{3}{5}$

Therefore, the probability that both numbers are prime would be $\frac{3}{5}$ .

4 0

2 years ago

At a grocery store the manager randomly selects 20 cartons of eggs, he finds 3 cartons with at least 1 broken egg. If the refrig

ivann1987 [24]

Answer:

You would expect 75 cartons with at least 1 broken egg.

There are likely to be 222 DVD players that are defective.

Step-by-step explanation:

Both examples given in this question represent a ratio of broken or defective to a sample number of cartons or DVD players. In order to find how many would be broken or defective overall, you can set up equivalent ratios, or proportions.

Egg Cartons: $\frac{broken}{#cartons}=\frac{3}{20}=\frac{x}{500}$

'x' represents the number of cartons out of 500 that would be expected to have at least 1 broken egg. To solve for 'x' you can cross-multiply and divide, or you can find what factor you would multiply the denominator by to get 500 and then multiply the same factor by the numerator to find 'x'. In this case 500/20 = 25, so the factor is 25. 3 x 25 = 75, so you have 75 expected cartons with a broken egg.

DVD Players: $\frac{defective}{#players}=\frac{3}{50}=\frac{x}{3700}$

'x' represents the number of DVD players out of 3700 that would be expected to be defective. To solve for 'x' you can cross-multiply and divide, or you can find what factor you would multiply the denominator by to get 3700 and then multiply the same factor by the numerator to find 'x'. In this case 3700/50 = 74, so the factor is 74. 3 x 74 = 222, so you have 22 likely defective DVD players.

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2 years ago