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sweet [91]
3 years ago
7

A claw arcade machine contains stuffed animals and mystery boxes. There are 51 items in the machine. Winning a stuffed animal ha

s twice as many favorable outcomes as winning a mystery box. How many of each item is in the machine?
Mathematics
1 answer:
Oliga [24]3 years ago
4 0
Answer: 34 stuffed animals & 17 mystery boxes.

First off, split 51 into three parts, trying to find how much 1/3 is of 51. It equals 17.

The stuffed animals in the machine are “twice as much” of the mystery boxes, so you need to multiply 17 x 2 to find the amount of stuffed animals.

17 x 2 = 34

As for the mystery boxes, you need to multiply it by one.

17 x 1 = 17.

In summary, the answer is 34 stuffed animals and 17 mystery boxes.
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Use the graph of the function to estimate the intervals on which the function is increasing or decreasing. (Enter your answers u
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<h3>Function</h3>

A function can be classified as increasing or decreasing. Thus, a function is increasing when the y-values increase, on the other hand, a function is decreasing when the y-values decrease.

From the image, you can see that the y-values increase in the following x-intervals: from -∞ to -3 and from 0.5 to ∞. Using interval notation, you can write that the function is increasing in:

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brainly.com/question/2649645

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Read 2 more answers
Lupe’s Construction is building a rectangular house that is 16 feet longer than it is wide. They will be putting a 3-foot wide s
mr Goodwill [35]

Answer:

The length of the four pieces of protective strip is 96 feet

Step-by-step explanation:

Let the w be the width of the rectangular house and L is length. Since L = w + 16, its perimeter P = 2(w + L)

= 2(w + w + 16)

= 2(2w + 16)

Since the perimeter of the house P = 136 feet,

136 = 2(2w + 16)

dividing through by 2, we have

136/2 = (2w + 16)

68 = (2w + 16)

collecting like terms, we have

68 - 16 = 2w

52 = 2w

dividing through by 2, we have

w = 52/2

w = 26 feet.

Now, since the side walk is 3 foot wide all around, the width of the house plus sidewalk = w + 3 and the length of the house plus sidewalk = L + 3 = w + 16 + 3 = w + 19.

So, the perimeter of the house plus sidewalk which is the perimeter of the four pieces of protective strip is P' = 2(w + 3 + w + 19)

= 2(w + 22)

= 2(26 + 22)

= 2(48)

= 96 feet

So, the length of the four pieces of protective strip is 96 feet.

5 0
4 years ago
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