All categories

3 years ago

In a company 70% of the workers are men. If 1590 people work for the company who arent men, how many workers are there in all?

Mathematics

2 answers:

Goryan [66]3 years ago

4 0

Answer: I believe the answer would be 5300.

meriva3 years ago

3 0

Answer:

5300

Step-by-step explanation:

1590 people are not men so they are women

$\frac{100 - 70}{100 } = \frac{1590}{x} \\ \\ \frac{3}{10} = \frac{1590}{x} \\ \\ x = 5300 \: totoal \: workers$

You might be interested in

There is large clock on the wall so student can check the time.Gym class starts at 1:15pm and end at 2:00 pm

lys-0071 [83]

It's 35 minutes passed

7 0

3 years ago

Read 2 more answers

Svetlana's hair is 4 centimeters long. Her hair grows 1.5 centimeters per month. Svetlana wants her hair to grow so that it is a

kap26 [50]

It would take her 2.6 months

7 0

3 years ago

If the measure of arc AB is 45 degrees and the radius is 12 meters, what is the arc length of arc AB?

dalvyx [7]

Answer:

9.42 m

Step-by-step explanation:

Arc AB = 2πr x (45/360)

= (2 x 3.14 x 12) / 8

= 9.42

3 0

3 years ago

An ant moves forward 18 1/2 inches in one hour. It turns around and crawlsbackward by 13.5 inches in the next hour. Finally, in

andrew-mc [135]

Answer:

Whats the question

Step-by-step explanation:

6 0

3 years ago

Solve the initial value problem

tigry1 [53]

$9(t+1)\dfrac{\mathrm dy}{\mathrm dt}-7y=14t\implies\dfrac{\mathrm dy}{\mathrm dt}-\dfrac7{9(t+1)}y=\dfrac{14t}{9(t+1)}$

Look for an integrating factor $\mu(t)$ :

$\ln\mu=\displaystyle-\frac79\int\frac{\mathrm dt}{t+1}=-\frac79\ln(t+1)\implies\mu=(t+1)^{-7/9}$

Multiply both sides by $\mu$ :

$(t+1)^{-7/9}\dfrac{\mathrm dy}{\mathrm dt}-\dfrac79(t+1)^{-16/9}y=\dfrac{14}9t(t+1)^{-16/9}$

Condense the left side as the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[(t+1)^{-7/9}y\right]=\dfrac{14}9t(t+1)^{-16/9}$

Integrate both sides:

$(t+1)^{-7/9}y=\displaystyle\frac{14}9\int t(t+1)^{-16/9}\,\mathrm dt$

For the integral on the right, substitute

$u=t+1\implies t=u-1\implies\mathrm dt=\mathrm du$

$\displaystyle\int t(t+1)^{-16/9}\,\mathrm dt=\int(u-1)u^{-16/9}\,\mathrm du$

$\displaystyle=\int\left(u^{-7/9}-u^{-16/9}\right)\,\mathrm du=\frac92u^{2/9}+\frac97u^{-7/9}+C$

$\implies(t+1)^{-7/9}y=\dfrac{14}9\left(\dfrac92(t+1)^{2/9}+\dfrac97(t+1)^{-7/9}+C\right)$

$\implies(t+1)^{-7/9}y=7(t+1)^{2/9}+2(t+1)^{-7/9}+C$

$\implies y=7(t+1)+2+C(t+1)^{7/9}=7t+9+C(t+1)^{7/9}$

Given that $y(0)=12$ , we get

$12=9+C\implies C=3$

$\implies\boxed{y(t)=7t+9+3(t+1)^{7/9}}$

6 0

3 years ago