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monitta
2 years ago
5

The annual 2-mile fun-run is a traditional fund-raising event to support local arts and sciences activities. It is known that th

e mean and the standard deviation of finish times for this event are respectively \mu μ = 30 and \sigma σ = 5.5 minutes. Suppose the distribution of finish times is approximately bell-shaped and symmetric. Find the approximate proportion of runners who finish in under 19 minutes.
Mathematics
1 answer:
victus00 [196]2 years ago
8 0

Answer: 0.0228

Step-by-step explanation:

Given : The  mean and the standard deviation of finish times (in minutes) for this event are respectively as :-

\mu=30\\\\\sigma=5.5

If the distribution of finish times is approximately bell-shaped and symmetric, then it must be normally distributed.

Let X be the random variable that represents the finish times for this event.

z score : z=\dfrac{x-\mu}{\sigma}

z=\dfrac{19-30}{5.5}=-2

Now, the probability of runners who finish in under 19 minutes by using standard normal distribution table :-

P(X

Hence, the approximate proportion of runners who finish in under 19 minutes = 0.0228

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4. Determine whether f(x) = 1/3x + 5 and g(x) = 3x - 15 are inverse functions.
Ierofanga [76]

Answer:

YES! we conclude that f(x) = 1/3x + 5 and g(x) = 3x - 15 are inverse functions.

Step-by-step explanation:

Given

Given that the function f(x) and g(x) are inverse functions.

f\left(x\right)\:=\:\frac{1}{3}x\:+\:5

g(x) = 3x - 15

To determine

Let us determine whether f(x) = 1/3x + 5 and g(x) = 3x - 15 are inverse functions.

<u>Determining the inverse function of f(x) </u>

A function g is the inverse function of f if for y = f(x), x = g(y)

y=\frac{1}{3}x+5

Replace x with y

x=\frac{1}{3}y+5

Solve for y

y=3x-15

Therefore,

YES! we conclude that f(x) = 1/3x + 5 and g(x) = 3x - 15 are inverse functions.

4 0
2 years ago
If the dimensions of a pentagonal prism are quadrupled, then the surface area of the prism is multiplied by eight.
Lubov Fominskaja [6]

Answer:

false

Step-by-step explanation:

the relationship between lengths/dimensions and areas is that areas are created by multiplying 2 dimensions.

when you quadruple (×4) the dimensions, then the areas are growing with the square of the factor (×4×4 = ×16), because the factor goes twice into the multiplication : one time for every dimension involved.

so, quadrupling the dimensions would multiply the areas by 16.

8 0
2 years ago
Robert accepted a new job at a company with a contract guaranteeing annual raises. Let S represent Robert's salary after working
vesna_86 [32]

Answer:

74,000

Step-by-step explanation:

In the year where he got 104,000 the year before he got 6,000 less. The year before 98,000 he got 12,000 less which means that now it would just be 86,000-12,000=74,000

I think this is right but you can double check on a calculator

Hope this helped!

6 0
2 years ago
How to solve the system of equation 500y-600x=2800 , 400y-400x=2400 by elimination
tatiyna

Answer:

x=2 and y=8

Step-by-step explanation:

Given:

Equation 1:

500y-600x=2800

Simplifying the above equation by dividing both sides by 100

\frac{500y-600x}{100}=\frac{2800}{100}

\frac{500y}{100}-\frac{600x}{100}=28\\\\5y-6x=28

Equation 2:

400y-400x=2400

Simplifying the above equation by dividing both sides by 400.

\frac{400y-400x}{400}=\frac{2400}{400}

\frac{400y}{400}-\frac{400x}{400}=6\\\\y-x=6

Now the system of equation is:

(1a) 5y-6x=28

(2a) y-x=6

Solving by elimination

Multiplying equation (2a) with -5

-5(y-x)=-5\times 6

(2b) -5y+5x=-30

Adding equations (1a) and (2b) in order to eliminate y

      5y-6x=28

+  -5y+5x=-30

We get -x=-2

∴ x=2

Plugging x=2 in equation (2a).

y-2=6

Adding 2 both sides.

y-2+2=6+2

∴ y=8

7 0
3 years ago
Noahzimmerman0987654321
sattari [20]

Answer:

ummm what is this for tho?

5 0
3 years ago
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