All categories

3 years ago

8.) Out of 600 employees, only 270 are happy with their pay. What percent is this? (Please show your work.) *

Mathematics

2 answers:

Gnom [1K]3 years ago

5 0

For this case we can raise a rule of three:

600 employees -------------> 100%

270 employees -------------> x

Where the variable "x" represents the percentage of employees who are satisfied with their salary. So, we have:

$x = \frac {270 * 100} {600}\\x = 45$

Thus, 45% of employees are satisfied with their salary.

Answer:

45%

andreev551 [17]3 years ago

5 0

ANSWER

$45\%$

EXPLANATION

The total number of employees is 600.

The number of employees who are happy with their pay is 270.

The percentage of employees who are happy with their pay is the number who are happy with their pay divided by total number of employees times 100%

$\frac{270}{600} \times 100\%$

This simplifies to

$45\%$

You might be interested in

The cost of a service or item is a(n): 1. Expense 2. Income 3. Coupon 4. Budget

miv72 [106K]

The cost of a service or item is an expense

5 0

3 years ago

I need help on how to get the answer too number 27.

KiRa [710]

A^2 + b^2 = c^2
3^2 + 8^2 = c^2
9 + 64 = c^2
73 =c^2
take the square root of each side
c= 8.544

3 0

3 years ago

Lily took a survey of 80 students on 7th grade and 150 students in 8th grade she found that 80% of the 7th graders like country

Anarel [89]

Answer:

Step-by-step explanation:

80 over 100*150

80 divided by 100=0.8 *150=1208student

0.8 *60=48

8 0

3 years ago

Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview

svetlana [45]

Answer: The required solution is

$y=(-2+t)e^{-5t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.$

So, the general solution of the given equation is

$y(t)=(A+Bt)e^{-5t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.$

According to the given conditions, we have

$y(0)=-2\\\\\Rightarrow A=-2$

and

$y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.$

Thus, the required solution is

$y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.$

6 0

3 years ago

If $120.99 is charged for 654 units of electricity

pochemuha

Answer:

The cost of one unit of electricity is $0.185

Step-by-step explanation:

Amount charged = $120.99

Units of electricity used = 654

The cost of a unit of electricity can be determined by dividing the amount charged by the number of units used.

So that,

cost of one unit of electricity = $\frac{120.99}{654}$

= 0.185

cost of one unit of electricity = $0.185

This implies that the cost of electricity used is charged at $0.185 per unit.

Thus for 654 units of electricity used, the cost would be;

654 x $0.185 = $120.99

7 0

3 years ago