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svetoff [14.1K]
3 years ago
7

What is the interquartile range of the following data set 2, 5, 9, 11, 18,30, 42, 55, 58, 73, 81

Mathematics
1 answer:
In-s [12.5K]3 years ago
4 0

Answer:

4.2/5. 4.

Step-by-step explanation:

You might be interested in
A board is 12 feet long. If the board is divided into 12 equal sections, how long is each section?
SIZIF [17.4K]
12 feet divided by 12 is equal to 1 foot ]
therefore a 12 foot long board divided into 12 equal sections has 12 1 foot long sections
6 0
3 years ago
Decompose 5/15 and 8/12
AnnZ [28]

Answer:

:)

Step-by-step explanation:

5/15 divided by 5 is equal to 1/3 and

8/12 divided by 2 is 4/6 divided by 2 is 2/3

You have to divide the numerator with the same number that you divide the denominator with.  

6 0
2 years ago
Pls help me with this
Anit [1.1K]

Answer:

hope this will help you more

8 0
2 years ago
Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10
zvonat [6]

The approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

<h3>What is depreciation?</h3>

Depreciation is to decrease in the value of a product in a period of time. This can be given as,

FV=P\left(1-\dfrac{r}{100}\right)^n

Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.

Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is n_1. Thus, by the above formula for the first car,

0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}

Take log both the sides as,

\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85

Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.

Thus, the depreciation price of the car is 0.6y. Let the number of year is n_2. Thus, by the above formula for the second car,

0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}

Take log both the sides as,

\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14

The difference in the ages of the two cars is,

d=4.85-3.14\\d=1.71\rm years

Thus, the approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

Learn more about the depreciation here;

brainly.com/question/25297296

4 0
2 years ago
C. Brian saves some money from his allowance to buy Mother a gift. His daily allowance is P50.00
Finger [1]

Answer:

1.thursday

his expenses was p26.85

2.monday he spent p40.8

3.18.90+23.15+11.80+16.35+9.20=approximately p55.6×30 days=p1,668

4.693.80-1,668

so he savings is completely enough

8 0
2 years ago
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