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Ostrovityanka [42]
2 years ago
5

What is the interquartile range (IQR) of this data set?

Mathematics
2 answers:
Andreas93 [3]2 years ago
7 0

Answer:

11 is the median of the entire data set

6 is the median of the lower half and 16 is the median of the upper half data scores...

Subtract 16 from 6 gives 10 so the answer is C) 10

Sever21 [200]2 years ago
6 0
<h3>Answer:  Choice C)  10</h3>

=============================================================

Explanation:

The first step is to sort the data from smallest to largest. Luckily, that's already been done for us.

Next, we find the median. This is the middle most number of the set. In this case, the median is 11 since we have 3 values below it (2,6,7) and 3 values above it (15,16,17).

Let

L = {2,6,7}

U = {15,16,17}

where L and U are the lower and upper sets respectively.

In other words, anything in set L is lower than the median while anything in set U is above the median. The median itself (11) is not part of either set.

Notice how 6 is the midpoint of set L, while 16 is the midpoint of set U.

This means Q1 = 6 and Q3 = 16

Therefore,

IQR = Q3 - Q1

IQR = 16 - 6

IQR = 10

which is why choice C is the final answer.

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Answer:

t=15.4\ years

Step-by-step explanation:

The  exponential growth function compounded continuously is equal to

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P is the initial population  

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t is Number of years

e is the mathematical constant number

we have  

A=4x\\P=x\\ r=9\%=9/100=0.09  

substitute in the function above

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Take natural log of both sides

ln(4)=ln[(e)^{0.09t}]

ln(4)=0.09t(ln(e))

ln(e)=1

ln(4)=0.09t

t=ln(4)/0.09

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(x-yi)(3+5i) is the conjugate of 6+24i
Nataly_w [17]

Answer:

x=-3

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(-3-3i)(3+5i)

Step-by-step explanation:

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5 0
2 years ago
PLEASE HELP DUE ON THURSDAY
valentinak56 [21]

Answer:

  1. 615.8 m^2
  2. 314.2 cm^2
  3. 38.5 ft^2
  4. 113.1 in^2
  5. 1075.2 m^2 . . . . corrected to 107,521.0 m^2
  6. 490.9 yd^2
  7. 18.8 m
  8. 94.2 in
  9. 81.7 ft
  10. 59.7 km
  11. 131.9 cm
  12. 144.5 yd

Step-by-step explanation:

The formula for the area of a circle is ...

   A = πr^2

For each of the given circles, use the value of the radius in this formula. For example, for problem 2, the diameter is shown as 20 cm, so the radius is half that, 10 cm. Putting that value into the formula gives ...

  A = 3.14159·(10 cm)^2 = 314.159 cm^2 ≈ 314.2 cm^2

___

The formula for the circumference of a circle is ...

  C = 2πr . . . . when the radius is given

or ...

  C = πd . . . . . when the diameter is given

As above, put the value for radius (or diameter) into the appropriate formula and evaluate it. For problem 12, that is ...

  C = 3.14159·(46 yd) = 144.513 yd ≈ 144.5 yd

_____

<em>Comment on doing lots of problems</em>

The idea behind having you do lots of problems is that repetition will help you remember the formula and/or how to use it. Personally, I find it tedious, so I like to let a spreadsheet or graphing calculator do the arithmetic when there are lots of identical problems (just with different numbers). This question is posted in the college math forum, so we assume you have a graphing calculator and know how to use it.

The attached photo shows a calculator set up to do the calculation and the rounding for each of the problem types. To get the answer to the last area calculation requires scrolling to the right to find it is 490.9 yd^2.

Note that we have used the calculator's button for π. If you put the number in yourself, use the one shown above, or the ratio 355/113, which is accurate to 7 digits.

In some cases, the numbers in the photo are not terribly clear. I took my best guess. In the event I read the value or units incorrectly, you will have to make appropriate adjustments—that is, do the calculation yourself.

_____

Late edit: apparently I read the number wrong for problem 5. The area of a circle with radius 185 m is 107,521.0086 m^2, which rounds to 107,521.0 m^2. Oops.

3 0
3 years ago
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