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lara [203]
2 years ago
11

Charlene had 5,750 pieces of penny candy. She sold 3/5 of them and divided rest into 5 equal bags to give her friends for Hallow

een. How many pieces of candy were in each bag? (PLEASE SHOW STEPS!)
Mathematics
1 answer:
Musya8 [376]2 years ago
6 0
(3/5)*5750=3450
3450÷5=690
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Dominik [7]
Answer: all of them
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2 years ago
It takes you 5 hours to complete a 375-kilometer trip. What was your average speed on this trip?
morpeh [17]

Answer:

75

Step-by-step explanation:

V=S/T(the formula)

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4 0
3 years ago
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A local gym instructor has a course load that allows her to teach eight classes. At an interest meeting, 8 people wanted high-­‐
cricket20 [7]

Answer:

<u>The final curse load of the local gym instructor is this:</u>

<u>High-­‐impact aerobics = 1</u>

<u>Low-­‐impact aerobics =  4</u>

<u>Jazzercise = 1</u>

<u>Step exercise = 2</u>

Step-by-step explanation:

1. Let's apportion the class load using the Hamilton method

Number of classes the local gym instructor can teach = 8

Total number of students that want to take a class = 114 (8 people wanted high-­‐impact aerobics, 64 wanted low-­‐impact aerobics, 11 wanted jazzercise, and 31 wanted step exercise)

Standard divisor = 114/8 = 14.25

Now, we can apportion the students in in Standard quotas, this way:

High-­‐impact aerobics = 8/14.25 = 0.5614

Low-­‐impact aerobics = 64/14.25 = 4.49

Jazzercise = 11/14.25 = 0.7719

Step exercise = 31/14.25 = 2.1754

Now, we find the Minimum quota, just considering the whole number and don't taking into account the decimals, this way:

High-­‐impact aerobics = 0

Low-­‐impact aerobics =  4

Jazzercise = 0

Step exercise = 2

As we can see we have 6 classes and there are 2 still pending. Those 2 goes to the classes with the highest decimal portion, in this case, Jazzercise .7719 and High-­‐impact aerobics with .5614.

<u>The final course load is this:</u>

<u>High-­‐impact aerobics = 1</u>

<u>Low-­‐impact aerobics =  4</u>

<u>Jazzercise = 1</u>

<u>Step exercise = 2</u>

7 0
3 years ago
Please dont let this go by.. I really need help
matrenka [14]

What do you need help with?

3 0
3 years ago
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Which equation in slope-intercept form represents a line that passes through the point (6,−1) and is perpendicular to the line y
damaskus [11]

For this case we have that by definition, the equation of a line of the slope-intersection form is given by:

y = mx + b

Where:

m: It is the slope of the line

b: It is the cut point with the y axis

By definition, if two lines are perpendicular then the product of their slopes is -1.

If we have: y = 2x-7

m_ {1} = 2\\2 * m 2 = - 1\\m_ {2} = - \frac {1} {2}

Thus, the equation is of the form:

y = - \frac {1} {2} x + b

We substitute the point:

-1 = - \frac {1} {2} (6) + b\\-1 = - \frac {1} {2} (6) + b\\-1 = -3 + b\\-1 + 3 = b\\b = 2

Finally, the equation is:

y = - \frac {1} {2} x + 2

Answer:

y = - \frac {1} {2} x + 2

6 0
2 years ago
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