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Anon25 [30]
2 years ago
7

9 friend catches the bus into town.the total cost is 31.05.how much does it cpst for one person

Mathematics
2 answers:
Kruka [31]2 years ago
7 0
I think the answer is 3.45$ for one person.
natulia [17]2 years ago
5 0
31.05 / 9 = 3.45....it cost each person $ 3.45
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The surface area of this cube is 96 square meters. what is the value of y?
ELEN [110]
The formula for finding the surface area of a cube is A=6y^2 where y is the side length. we know a=96. so we just isolate y.
96/6= 6y^2 / 6
16=y^2
sqrt (16) = sqrt(y^2)
4=y

So y is 4

5 0
3 years ago
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What is the answer to 2a + 3b =
Rainbow [258]

Answer:

2a+3b

Step-by-step explanation:

There wouldn't be a simplfied answer or solution if you don't have a solution for either a or b or both. And they are not like terms, so you can't add them together.

So the answer is 2a + 3b.

I hope this helps!

Have a great day!

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2 years ago
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John's friend told him that he could earn $49 for handing out flyers at a local concert. John wants to calculate the hourly rate
gladu [14]
X= $14
In other words you can do 49 divided by 3.5
6 0
3 years ago
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Factor 70q+60. Write your answer as a product with a whole number greater than 1.
faltersainse [42]

Answer:

10(7q + 6)

Step-by-step explanation:

  1. Find the GCF of 70 and 60: 10
  2. Divide each number by 10 (steps shown below)
  3. 70q ÷ 10 = 7q
  4. 60 ÷ 10 = 6
  5. Re-write the expression with 10: 10(7q + 6)

I hope this helps!

7 0
3 years ago
The mean percentange of a population of people eating out at least once a week is 57℅
Sidana [21]

Answer:

<u>The correct answer is B. between 56.45% and 57.55% </u>

Complete statement and question:

The mean percentage of a population of people eating out at least once a week is 57% with a standard deviation of 3.50%. Assume that a sample size of 40 people was surveyed from the population a significant number of times. In which interval will 68% of the sample means occur?

between 55.89% and 58.11%

between 56.45% and 57.55%

between 56.54% and 57.46%

between 56.07% and 57.93%

Source: brainly.com/question/1068489

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

Mean percentage of a population of people eating out at least once a week  = 57%

Standard deviation = 3.5%

Sample size = 40

Confidence level = 68%

2. In which interval will 68% of the sample means occur?

For answering this question, we should find out the standard deviation of the sample, using this formula:

Standard deviation of the sample = Standard deviation of the population/√Sample size

Standard deviation of the sample = 3.5/√40

Standard deviation of the sample = 3.5/6.32

Standard deviation of the sample = 0.55

Let's recall that a confidence level of 68% means that 68% of the sample data would have a value between the mean - 1 time the standard deviation of the sample and the mean  + 1 time the standard deviation of the sample. Thus:

57 - 1 * 0.55 = 57 - 0.55 = 56.45

57 + 1  * 0.55 = 57 + 0.55 = 57.55

<u>The correct answer is B. between 56.45% and 57.55% </u>

7 0
3 years ago
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