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Bezzdna [24]
3 years ago
9

There are 12 boys and 8 girls in a class. If 2 more girls join the class, what is the new ratio for the number of boys to the nu

mber of girls in the class?
Mathematics
1 answer:
ycow [4]3 years ago
4 0
Here how it goes, 12 boys to 8 girls is 3:2 in ratio form but since u adding 2 more girls then 8 girls + 2 girls = 10 girls so now its 12 boys + 10 girls so your answer in final would be 6:5.
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Answer:

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Step-by-step explanation:

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2 years ago
Thirty-five percent of what number is 17.5?
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4 0
3 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
Robert climbed 775 steps in 12 1/2 minutes. How many steps did he average per minute
Archy [21]
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6 0
3 years ago
What is the value of x?
Alex_Xolod [135]

Answer:

x=4

Step-by-step explanation:

ABCD is a rectangle, so AD and BC are parallel. Thus, The angle complementary to 6x+19 is 47.

6x+19+47=90

6x+66=90

6x=24

x=4

7 0
3 years ago
Read 2 more answers
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