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Nookie1986 [14]
1 year ago
8

Easton wants to put a variety of donuts in a display case. The case is large enough to hold 100 donuts. Complete the table below

.

Mathematics
1 answer:
shutvik [7]1 year ago
3 0

The amounts for each donut are given as follows:

  • Strawberry: 15.
  • Chocolate: 45.
  • Sprinkle: 40.

<h3>How to obtain the amounts?</h3>

The amounts for each donut are found applying the proportion, multiplying the decimal equivalent of each percentage by the total number of donuts.

From the problem, the total number of donuts is given as follows:

100 donuts.

The percentages of each type of donuts are given as follows:

  • Strawberry: 15%.
  • Chocolate: 45%.
  • Sprinkle: 40%.

The decimal equivalent of each percentage is given as follows:

  • Strawberry: 0.15.
  • Chocolate: 0.45.
  • Sprinkle: 0.4.

Hence the amounts are given as follows:

  • Strawberry: 15, as 0.15 x 100 = 15.
  • Chocolate: 45, as 0.45 x 100 = 45.
  • Sprinkle: 40, as 0.4 x 100 = 40.

More can be learned about proportions at brainly.com/question/24372153

#SPJ1

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Step-by-step explanation:

Nadia's is 41 years old.

8 0
3 years ago
The volume of a right circular cone varies jointly as the altitude and the square of the radius of the base. If the volume of th
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Answer:13.5 inches

Step-by-step explanation:

Here, the volume of a right circular cone varies jointly as the altitude and the square of the radius of the base

So our equation for volume becomes

V = c*h*r^2

where 'h' is altitude and 'r' is radius of base and 'c' is constant

Putting the value of V,h,r,

we get,

154 = c*12*3.5*3.5

c = 22/21

Now we have volume = 77 cu and radius of the base =7/3, so putting the values we get,

77 = 22/21*h*7/3*7/3

or, h= 13.5

Hope it helps!!!

6 0
3 years ago
Read 2 more answers
4) Round 694,553 to the nearest hundred thousand. O 600,000 O 694,600 O 695,000 O 700,000​
Rudiy27

Answer:

700,000

Step-by-step explanation:

8 0
3 years ago
A fruit company delivers its fruit in two types of boxes: large and small. A delivery of
Tema [17]

Answer:

Small box weighs 13.75 kg & large box weighs 15.75 kg

Step-by-step explanation:

We can write 2 simultaneous equation and solve for weight of each box.

<em>Let weight of large box be l and small box be s.</em>

<em />

"<u>3 large boxes and  5 small boxes has a total weight of  116 kilograms</u>":

3l+5s=116

and

"<u>9 large boxes and  7 small boxes has a total weight of  238 kilograms</u>":

9l+7s=238

<em>Now we can solve for l in the 1st equation and put it into 2nd equation and get s:</em>

<em>3l+5s=116\\3l=116-5s\\l=\frac{116-5s}{3}</em>

<em>now,</em>

<em>9l+7s=238\\9(\frac{116-5s}{3})+7s=238\\3(116-5s)+7s=238\\348-15s+7s=238\\348-238=15s-7s\\110=8s\\s=\frac{110}{8}=13.75</em>

<em />

<em>now we plug in 13.75 into s into equation of l to find s:</em>

<em>l=\frac{116-5s}{3}\\l=\frac{116-5(13.75)}{3}\\l=15.75</em>

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3 years ago
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