All categories

1 year ago

The histogram below shows the estimated monthly salaries of company employees with different years of experience.

Mathematics

2 answers:

skad [1K]1 year ago

8 0

From the Histogram displaying the the estimated monthly salaries of company employees with different years of experience, the graph might be misleading given that:
The scale on the y-axis overemphasizes the difference in salaries between different experience levels.

It might not be the case always that salary increment is proportional to experience, this might not fit all the cases.

daser333 [38]1 year ago

4 0

The correct answer is A. The scale on the y-axis overemphasizes the difference in salaries between different experience levels.

You might be interested in

19. Bella is putting down patches of sod

Fofino [41]

Answer:

The dimensions of the two different rectangular regions are;

1st Arrangement:

W = 4 yards and L = 5 yards or W = 5 yards and L = 4 yards

2nd Arrangement:

W = 2 yards and L = 10 yards or W = 10 yards and L = 2 yards

The perimeter of the two different rectangular regions are;

1st Arrangement:

P₁ = 18 yards

2nd Arrangement:

P₂ = 24 yards

Step-by-step explanation:

Bella is putting down patches of sod to start a new lawn.

She has 20 square yards of sod.

We are asked to provide the dimensions of two different rectangular regions that she can cover with the sod.

Recall that a rectangle has an area given by

Area = W*L

Where W is the width of the rectangle and and L is the length of the rectangle.

Since Bella has 20 square yards of sod,

20 = W*L

There are more than two such possible rectangular arrangements.

Out of them, two different possible arrangements are;

1st Arrangement:

20 = (4)*(5) = (5)*(4)

Width is 4 yards and length is 5 yards or width is 5 yards and length is 4 yards

2nd Arrangement:

20 = (2)*(10) = (10)*(2)

Width is 2 yards and length is 10 yards or width is 10 yards and length is 2 yards

Therefore, the dimensions of two different rectangular regions are;

1st Arrangement:

W = 4 yards and L = 5 yards or W = 5 yards and L = 4 yards

2nd Arrangement:

W = 2 yards and L = 10 yards or W = 10 yards and L = 2 yards

What is the perimeter of each region?

The perimeter of a rectangular shape is given by

P = 2(W + L)

Where W is the width of the rectangle and and L is the length of the rectangle.

The perimeter of the 1st arrangement is

P₁ = 2(4 + 5)

P₁ = 2(9)

P₁ = 18 yards

The perimeter of the 2nd arrangement is

P₂ = 2(2 + 10)

P₂ = 2(12)

P₂ = 24 yards

So the perimeter of the 1st arrangement is 18 yards and the perimeter of the 2nd arrangement is 24 yards.

Note:

Another possible arrangement is,

20 = (1)*(20) = (20)*(1)

Width is 1 yard and length is 20 yards or width is 20 yards and length is 1 yard.

3 0

1 year ago

A large tank is partially filled with 100 gallons of fluid in which 20 pounds of salt is dissolved. Brine containing 1 2 pound o

Valentin [98]

Answer:

47.25 pounds

Step-by-step explanation:

$\dfrac{dA}{dt}=R_{in}-R_{out}$

First, we determine the Rate In

Rate In=(concentration of salt in inflow)(input rate of brine)

$=(0.5\frac{lbs}{gal})( 6\frac{gal}{min})\\R_{in}=3\frac{lbs}{min}$

Change In Volume of the tank, $\frac{dV}{dt}=6\frac{gal}{min}-4\frac{gal}{min}=2\frac{gal}{min}$

Therefore, after t minutes, the volume of fluid in the tank will be: 100+2t

Rate Out

Rate Out=(concentration of salt in outflow)(output rate of brine)

$R_{out}=(\frac{A(t)}{100+2t})( 4\frac{gal}{min})\\\\R_{out}=\frac{4A(t)}{100+2t}$

Therefore:

$\dfrac{dA}{dt}=3-\dfrac{4A(t)}{100+2t}\\\\\dfrac{dA}{dt}=3-\dfrac{4A(t)}{2(50+t)}\\\\\dfrac{dA}{dt}=3-\dfrac{2A(t)}{50+t}\\\\\dfrac{dA}{dt}+\dfrac{2A(t)}{50+t}=3$

This is a linear differential equation in standard form, therefore the integrating factor:

$e^{\int \frac{2}{50+t}dt}=e^{2\ln|50+t|}=e^{\ln(50+t)^2}=(50+t)^2$

Multiplying the DE by the integrating factor, we have:

$(50+t)^2\dfrac{dA}{dt}+(50+t)^2\dfrac{2A(t)}{50+t}=3(50+t)^2\\\{(50+t)^2A(t)\}'=3(50+t)^2\\$Taking the integral of both sides\\\int \{(50+t)^2A(t)\}'= \int 3(50+t)^2\\(50+t)^2A(t)=(50+t)^3+C $ (C a constant of integration)\\Therefore:\\A(t)=(50+t)+C(50+t)^{-2}$

Initially, 20 pounds of salt was dissolved in the tank, therefore: A(0)=20

$20=(50+0)+C(50+0)^{-2}\\20-50=C(50)^{-2}\\C=-\dfrac{30}{(50)^{-2}} =-30X50^2=-75000$

Therefore, the amount of salt in the tank at any time t is:

$A(t)=(50+t)-75000(50+t)^{-2}$

After 15 minutes, the amount of salt in the tank is:

$A(15)=(50+15)-75000(50+15)^{-2}\\=47.25$ pounds$

8 0

1 year ago

Find the value of $x$ that maximizes $$f(x) = \log (-20x + 12\sqrt{x}).$$ If there is no maximum value, write "NONE".

kakasveta [241]

The function is written as:
f(x) = log(-20x + 12√x)
To find the maximum value, differentiate the equation in terms of x, then equate it to zero. The solution is as follows.

The formula for differentiation would be:
d(log u)/dx = du/u ln(10)
Thus,
d/dx = (-20 + 6/√x)/(-20x + 12√x)(ln 10) = 0
-20 + 6/√x = 0
6/√x = 20
x = (6/20)² = 9/100

Thus,
f(x) = log(-20(9/100)+ 12√(9/100)) = 0.2553

The maximum value of the function is 0.2553.

5 0

2 years ago

A scientist who studies teenage behavior was interested in determining if teenagers spend more time playing computer games then

Marina86 [1]

Answer:

- The amount of time greater than 10.2 hours for this year.

- The $P_{value}$ is 0.01358.