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harkovskaia [24]
3 years ago
11

SOMEONE PLS HELP ASAP

Mathematics
2 answers:
Viefleur [7K]3 years ago
8 0

Answer:

The answers are as follows:

After reading and comparing the values, we get the following answer.

Number of students with blue eyes blond hair  are 42.

Difference of the number of students with gray eyes and brown hair and number of students with green eyes and black hair is 11-5=6

Number of students with gray eyes brown hair  are 5.

Difference of the number of students with green eyes and blond hair and the number of students with gray hair is 35-21=14 because only gray hair marginal total is 35.

harkovskaia [24]3 years ago
5 0
42---number of students with blue eyes blond hair
5--students with grey eyes brown hair 
6--difference of the number of students with grey eyes and brown hair ...
14-- with the last one standing
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Ineed help help me please
melamori03 [73]
The symbol ₁₂P₉ represents the permutations of 9  quantities out of 12.
By definition,
_{12}P_{9} =  \frac{12!}{(12-9)!} = \frac{12!}{3!}

From the calculator,
12! = 479,001,600
3! = 6

Therefore
₁₂P₉ = 479001600/6 = 79,833,600

Answer: 79,833,600
8 0
3 years ago
Think about your previous math courses. Identify one tope that you learned about which
saw5 [17]

The answer for this question I don’t think could be answered by anyone but yourself. To help you figure this out, just simply think about all of the math courses you’ve had previous to your current course and think about which course you did best in.

8 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
A traingle has a perimeter of 22 centimeters with a side of 7 centimeters with a side that is 7 centimeters long. the lengths of
Leokris [45]
The shortest side of the triangle is 10cm
3 0
3 years ago
Type the correct answer in each box. If necessary, use / for the fraction bar(s).
Tom [10]

Given:

The values are a=0.\overline{6},b=0.75.

To find:

The values of ab and \dfrac{a}{b}.

Solution:

We have,

a=0.\overline{6}

a=0.666...                     ...(i)

Multiply both sides by 10.

10a=6.666...                 ...(ii)

Subtracting (i) from (ii), we get

10a-a=6.666...-0.666...

9a=6

a=\dfrac{6}{9}

a=\dfrac{2}{3}

And,

b=0.75

b=\dfrac{75}{100}

b=\dfrac{3}{4}

Now, the product of a and b is:

ab=\dfrac{2}{3}\times \dfrac{3}{4}

ab=\dfrac{1}{2}

The quotient of a and b is:

\dfrac{a}{b}=\dfrac{\dfrac{2}{3}}{\dfrac{3}{4}}

\dfrac{a}{b}=\dfrac{2}{3}\times \dfrac{4}{3}

\dfrac{a}{b}=\dfrac{8}{9}

Therefore, the required values are ab=\dfrac{1}{2} and \dfrac{a}{b}=\dfrac{8}{9}.

6 0
3 years ago
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