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2 years ago

Five times my age 4 years ago is the same as 3 times my age in 2 years

Mathematics

1 answer:

GREYUIT [131]2 years ago

7 0

Your current age is 13 years.

<h3>How to write expression?</h3>

Five times my age 4 years ago is the same as 3 times my age in 2 years.

let

x = current age.

Therefore, the expression can be represented as follows

5(x - 4) = 3(x + 2)

open the bracket

5(x - 4) = 3(x + 2)

5x - 20 = 3x + 6

subtract 3x from both sides

5x - 3x - 20 = 3x - 3x + 6

2x - 20 = 6

add 20 to both sides

2x - 20 + 20 = 6 + 20

2x = 26

divide both sides by 2

2x / 2 = 26 / 2

x = 13

Therefore, your current age is 13 years.

learn more on expression here: brainly.com/question/4715724

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Sam ran 63,756 feet in 70 minutes. What is Sam’s rate in miles per hour?

MArishka [77]

Answer:

63756/70=910.8 ft per minute

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3 years ago

6 numbers are shown

Sonja [21]

Answer:

$-1\frac{1}{2}$ , $-\frac{3}{4}$ , $-\frac{37}{50}$ , $\frac{5}{4}$ , $1\frac{1}{4}$ , $2\frac{3}{4}$

Step-by-step explanation:

$1\frac{1}{4}$ , $-1\frac{1}{2}$ , $-\frac{37}{50}$ , $-\frac{3}{4}$ , $2\frac{3}{4}$ , $\frac{5}{4}$

Order everything, negative to the left side and positive to the right side

<u><em>Negative numbers:</em></u>

$-1\frac{1}{2}$ , $-\frac{37}{50}$ , $-\frac{3}{4}$

<u><em>Positive numbers:</em></u>

$1\frac{1}{4}$ , $2\frac{3}{4}$ , $\frac{5}{4}$

<em><u>Now order it on a number line:</u></em>

$-1\frac{1}{2}$ , $-\frac{3}{4}$ , $-\frac{37}{50}$ , $\frac{5}{4}$ , $1\frac{1}{4}$ , $2\frac{3}{4}$

8 0

3 years ago

If you shift the linear parent function, (X) = x down 6 units, what is the

jeka94

Answer:

Step-by-step explanation:

For a parent function y = f(x) and n > 0:

f(x) + n : move the graph n units up

f(x) - n : move the graph n units down

f(x + n) : move the graph n units to the left

f(x - n) : move the graph n units to the right

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

We have

f(x) = x

Transformation: 6 units down

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4 years ago

The probability that an American CEO can transact business in a foreign language is .20. Twelve American CEOs are chosen at rand

NNADVOKAT [17]

Answer with Step-by-step explanation:

We are given that

The probability that an American CEO can transact business in foreign language=0.20

The probability than an American CEO can not transact business in foreign language= $1-0.20=0.80$

Total number of American CEOs chosen=12

a. The probability that none can transact business in a foreign language= $12C_0(0.20)^0(0.80)^{12}$

Using binomial theorem $nC_r(1-p)^{n-r}p^r$

The probability that none can transact business in a foreign language= $\frac{12!}{0!(12-0)!}(0.8)^{12}=(0.8)^{12}$

b.The probability that at least two can transact business in a foreign language= $1-P(x=0)-p(x=1)=1-((0.8)^{12}+12C_1(0.8)^{11}(0.2))=1-((0.8)^{12}+12(0.8)^{11}}(0.2))$

c.The probability that all 12 can transact business in a foreign language= $12C_{12}(0.8)^0(0.2)^{12}$

The probability that all 12 can transact business in a foreign language= $\frac{12!}{12!}(0.2)^{12}=(0.2)^{12}$

5 0

3 years ago

I need help with this!!!

timofeeve [1]

just use A=BxH

Base times height that should give you the correct answer

6 0

3 years ago