Solve the equation using the Properties of Equality.

4. The annual salaries of employees in a large company are approximately

svetoff [14.1K]

a. What percent of people earn less than $40000?

Solution: Let S be the random variable of a salary of employee (in $), S ~ N(50000,20000). Then the random

variable X =−50000

20000

~N(0,1).

( < 40000) = ( <

40000 − 50000

20000 ) = ( < −0.5) = (−0.5) = 0.3085375.

Here Φ(x) denotes the cumulative distribution function of a standard normal distribution.

Answer: 31%.

b. What percent of people earn between $45000 and $65000?

Solution:

(45000 < < 65000) = (

45000 − 50000

20000 < <

65000 − 50000

20000 ) = (−0.25 < < 0.75)

= (0.75) − (−0.25) = 0.7733726 − 0.4012937 = 0.3720789.

Answer: 37%.

c. What percent of people earn more than $70000?

Solution:

( > 70000) = ( >

70000 − 50000

20000 ) = ( > 1) = 0.8413447.

Answer: 84%.

5 0

3 years ago

What is the surface area of the cylinder?

guajiro [1.7K]

Answer:

i think it is C 80π ft^2

that is because 10*8=80

6 0

3 years ago

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Evaluate 5 + 6 · 2 – 8 ÷ 4 + 7 using the correct order of operations.

allsm [11]

Answer:

22

Step-by-step explanation:

5+12-2+7

17-2+7

15+7

22

6 0

3 years ago

The average weight of the class of 35 students was 45 KG. with the admission of a new student the average weight came down to 44

melamori03 [73]

The weight of the new student is 27 kg.

Average weight

= total weight ÷total number of students

<h3>1) Define variables</h3>

Let the total weight of the 35 students be y kg and the weight of the new student be x kg.

<h3>2) Find the total weight of the 35 students</h3>

y= 35(45)

y= 1575 kg

<h3>3) Write an expression for average weight of students after the addition of the new student</h3>

New total number of students

= 35 +1

= 36

Total weight

= total weight of 35 students +weight of new students

= y +x

$44.5 = \frac{y + x}{36}$

<h3>4) Substitute the value of y</h3>

$44.5 = \frac{1575 + x}{36}$

<h3>5) Solve for x</h3>

36(44.5)= 1575 +x

1602= x +1575

<em>Subtract 1575 from both sides:</em>

x= 1602 -1575

x= 27

Thus, the weight of the new student is 27 kg.

7 0

3 years ago

A pencil consists of a cone stacked on top of a cylinder. The diameter of the cylindrical base of the pencil is 10 mm and the he

11Alexandr11 [23.1K]

Answer:

790π

Step-by-step explanation:

We are given;

Diameter of cylinder;d = 10 mm

So, radius;r = 10/2 = 5 mm

Height of cylinder;h = 70mm

Surface area of cylinder is given by the formula; S.A = 2πr² + 2πrh

Plugging in the relevant values, we have;S.A = 2π(5)² + 2π(5)(70)

S.A = 50π + 700π

S.A = 750π

Now, because one base of the cylinder is hidden as the cone is stacked on that face, we will deduct the area of that base face;

Thus, Surface area = 750π - π(5)² = 750π - 25π = 725π

For the cone,

Height;h = 12mm

Since this is stacked directly on the cylinder, it will have the same radius. Thus; radius;r = 5mm

Now,formula for surface area of cone is;

S.A = πr² + πrL

Where L is slant height.

We can use pythagoras theorem to get L.

So, L² = r² + h²

L = √r² + h²

L = √(5² + 12²)

L = √(25 + 144)

L = √169

L = 13

So, S.A of cone = π(5)² + (π×5×13)

S.A = 25π + 65π = 90π

Similar to what was done to the Cylinder, since the circular base of the cone is stacked on the cylinder, we will deduct the surface area of that base as it is hidden.

So, S.A is now = 90π - π(5)²

= 90π - 25π = 65π

Thus,total surface area of the pencil = 725π + 65π = 790π

7 0

3 years ago

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