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kondor19780726 [428]
3 years ago
15

The box plot represents the number of minutes customers spend on hold when calling a company. A number line goes from 0 to 10. T

he whiskers range from 2 to 8, and the box ranges from 3 to 6. A line divides the box at 5. What is the upper quartile of the data?
Mathematics
2 answers:
Sindrei [870]3 years ago
7 0

Answer:

6

Step-by-step explanation:

The upper quartile (also known as Q3) is the right most side of the BOX (where the box ends). In the question, it says the box is from 3 to 6 meaning that the box ends at 6.

I hope this helps!

zzz [600]3 years ago
5 0

Answer:

6

Step-by-step explanation:

The upper quartile is 6 because the box ends at 6 and that is the upper quartile which is sometimes also known as Q3

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Hi...I am taking pre calculus and I'm reallyo lost here..can someone help me?
Karolina [17]
1. Log7 (49)=2
2. Log63(216)=3
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3 0
4 years ago
Ben is 4 times as old as Ishaan and is also 6 years older than Ishaan. How old is Ishaan?
Ratling [72]
6 times 4 equals 24 that is the answer for this problem because it says that Ben is 4 times older than ishaan and he is 6
6 0
3 years ago
Read 2 more answers
Can someone help me for this question please? I just keep getting wrong answer! Pls someone explain to me step by step! ASAP!!!
Kobotan [32]

Answer:

  84 feet

Step-by-step explanation:

This problem involves several steps. The first step is to realize that the given figure does not show the required number of vertical stringers. It shows 5, but there will be 7 of them. The given diagram is helpful in that it shows a vertical stringer on the centerline of the arches.

The second step is to write a function that will tell you how long the stringer will be. I find it convenient to write the equation for an arch shape such as this using the parent function h(x) = 1-x^2. This parent function gives an arch of height 1 and width 1 from center (a total width of 2). You want an arch that is 16 ft high and 40 ft wide (one side from center), so you must scale this parent function both horizontally (by 40) and vertically (by 16). It becomes ...

  H(x) = 16(1 -(x/40)^2)

The taller arch is twice this height, so the length of a vertical stringer at position x is

  vertical length = 2H(x) -H(x) = H(x)

That is, the function H(x) we defined can be used to find the length of the stringers.

The third step is to find the stringer lengths. It can save some energy if you realize that the problem is symmetrical, so that the stringer at x=-30 is the same length as the one at x=30. We need to find stringer lengths every 10 feet from -40 feet to +40 feet. Of course, the ones at ±40 feet are zero length, because that is where the two arches meet.

  H(-30) = 16(1 -(3/4)^2) = 7

  H(-20) = 16(1 -(1/2)^2) = 12

  H(-10) = 16(1 -(1/4)^2) = 15

  H(0) = 16

Then the fourth step is to add up the stringer lengths, rounding the result as required.

  7 +12 +15 +16 +15 +12 +7 = 16 + 2(34) = 84 . . . . feet (no rounding needed)

Finally, you need to answer the question asked:

  The sum of vertical stringer lengths is 84 feet.

5 0
3 years ago
Can somebody plz help answer this word problem using equations thx :3 (only if u done this before) lol :)
ycow [4]

Answer:

one num = 6

another = 24

Step-by-step explanation:

One number = x

Another = 4x

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8 0
3 years ago
Round 128.59943 to 3 decimal places
Bond [772]
To 3 decimal places.

128.59943

Count three places to the right after the decimal point. The third digit is 9, and the fourth is 4. Since the 4th is not up to 5, we leave the 9 that way and so we can not round up the 9 to 10.

So it becomes = 128.599

Thanks.
7 0
4 years ago
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