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Alona [7]
3 years ago
6

What is the interquartile range of the data?   

Mathematics
2 answers:
Juli2301 [7.4K]3 years ago
7 0

Answer:

the answer is 26

amosc @briahall0

Step-by-step explanation:

zysi [14]3 years ago
3 0
An interquartile range is finding the median of the first half of the data set, and the second half, then subtracting the first median from the second one. 

The median of the data set is..

<span>17, 18, 18, 20, 23, 25, 27, 27, 34, 38, 40, 46, 46, 62, 67 

Now we have to find the interquartile points.

</span>17, 18, 18, 20, 23, 25, 27, 27, 34, 38, 40, 46, 46, 62, 67 

The interquartile points are 20 and 46. So subtract:

46 - 20 = 26

The answer is 26. :)

Hopefully this helps! If you have any more questions or don't understand, feel free to DM me, and I'll get back to you ASAP! :)
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Need help, extra points
castortr0y [4]

Answer:

around 35-42° it's one of them 2 probably 40-41°

4 0
3 years ago
2.The mean area of several thousand apartments in a new development is advertised to be 1,100 square feet. A consumer advocate h
KiRa [710]

Answer:

Null Hypothesis, H_0 : \mu = 1,100 square feet

Alternate Hypothesis, H_A : \mu < 1,100 square feet

Step-by-step explanation:

We are given that the mean area of several thousand apartments in a new development is advertised to be 1,100 square feet.

A consumer advocate has received numerous complaints that the apartments are smaller than advertised.

<u><em>Let </em></u>\mu<u><em> = mean area of several apartments.</em></u>

So, Null Hypothesis, H_0 : \mu = 1,100 square feet

Alternate Hypothesis, H_A : \mu < 1,100 square feet

Here, <u>null hypothesis states that</u> the mean area of apartments are same as advertised.

On the other hand, <u>alternate hypothesis states that</u> the mean area of apartments are smaller than advertised.

So, this would be the appropriate null and the alternative hypothesis to test this claim.

4 0
3 years ago
Sibal started with $500 in a bank account that does not earn interest. In the middle of every month, she withdraws 15 of the acc
Otrada [13]
After every month's withdrawals, 4/5 of the original amount at the start of the month will remain. The amount at the start of every month will change. Thus:
an = 4/5 (an-1) ; where a1 = 500.
4 0
3 years ago
An aquarium with a square base has no top. There is a metal frame. Glass costs 8 dollars/m^2 and the frame costs 7 dollars/m. Th
Irina-Kira [14]
Given that the volume of the aquarium is 20m^3.

Volume = Area of Base x height

Area of Base = Volume / height = 20/h

Given that the aquarium has a square base.

Area of square = l^2

Thus, the length of the base of the aquarium is \sqrt{area \ of \ base} = \sqrt{ \frac{20}{h} }

The frame is to cover 8 sides with the length equal to the length of the base and 4 sides with the length of the height.

Thus, the total perimeter of the frame is given by 8\sqrt{\frac{20}{h}}+4h= \sqrt{64\left(\frac{20}{h}\right)}+4h = \sqrt{\frac{1,280}{h}}+4h

Area of the four side faces of the aquarium is 4 times the length of the base times the height = 4\times\sqrt{ \frac{20}{h} }\times h=\sqrt{16\left(\frac{20}{h}\right)h^2}=\sqrt{320h}

Total area to be covered by grass is the base and the four side faces and is given by \frac{20}{h}+\sqrt{320h}

Cost of the entire metal frame = 7\left(\sqrt{\frac{1,280}{h}}+4h\right)= \sqrt{49\left(\frac{1,280}{h}\right)}+7(4h) = \sqrt{\frac{62,720}{h}}+28h

Cost of the entire grass = 8\left(\frac{20}{h}+\sqrt{320h}\right)=\frac{160}{h}+\sqrt{64(320h)}=\frac{160}{h}+\sqrt{20,480h

Therefore, total cost in terms of the height, h, is given by

C=\sqrt{\frac{62,720}{h}}+28h+\frac{160}{h}+\sqrt{20,480h
3 0
3 years ago
Transform the literal equation to solve for f.<br><br><br> fg + c = d
wolverine [178]

Answer:

f=(d-c)/g

Step-by-step explanation:

fg+c=d

fg=d-c

f=(d-c)/g

4 0
3 years ago
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