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Dovator [93]
3 years ago
5

Suppose the function, g(x), is used to model the height ,y, of a soccer ball, x seconds after the ball is kicked up in the air.

The ball starts on the ground and travels in a parabolic shape as it reaches a maximum height and then returns to the ground. Suppose further that the ball reaches its maximum height of 15 feet in 2.2 seconds. What would be an appropriate domain for g(x)?
Mathematics
1 answer:
andrey2020 [161]3 years ago
3 0

the ball reaches its maximum height of 15 feet in 2.2 seconds

The ball starts on the ground and travels in a parabolic shape as it reaches a maximum height and then returns to the ground

The ball starts on the ground at 0 seconds so the ball starts at x=0 seconds

the ball reaches its maximum height of 15 feet in 2.2 seconds

So it takes 2.2 second to reach the maximum height

It will take another 2.2 seconds to reach the ground

The time the ball in the air is 2.2 + 2.2 = 4.4

The ball will reach the ground in 4.4 seconds

So the domain for g(x) is {0,4.4}


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You consider yourself a bit of an expert at playing rock-paper-scissors and estimate that the probability that you win any given
cestrela7 [59]

Answer:

a) 0.37

b) 0.421

c) 0.25

Step-by-step explanation:

Since the probability of winning a game is binomial (P = 0.5) the expected value for number of winning when you play 60 games is

\mu = 60*0.5 = 30

And the standard deviation:

\sigma = 60*0.5*0.5 = 15

a) the cumulative probability of winning at least 35 games is

P(X \geq 35) = 1 - P(X < 35) = 1 - 0.631 = 0.37

b)P(X < 27) = 0.421

c)P(30 \leq X \leq 40) = P(X \leq 40) - P(X \leq 30) = 0.75 - 0.5 = 0.25

7 0
3 years ago
A manufacturer produces bearings, but because of variability in the production process, not all of the bearings have the same di
Lena [83]

Answer:

Proportion of all bearings falls in the acceptable range = 0.9973 or 99.73% .

Step-by-step explanation:

We are given that the diameters have a normal distribution with a mean of 1.3 centimeters (cm) and a standard deviation of 0.01 cm i.e.;

Mean, \mu = 1.3 cm            and           Standard deviation, \sigma = 0.01 cm

Also, since distribution is normal;

                 Z = \frac{X -\mu}{\sigma} ~ N(0,1)

Let X = range of diameters

So, P(1.27 < X < 1.33) = P(X < 1.33) - P(X <=1.27)

  P(X < 1.33) = P( \frac{X -\mu}{\sigma} < \frac{1.33 -1.3}{0.01} ) = P(Z < 3) = 0.99865

  P(X <= 1.27) = P( \frac{X -\mu}{\sigma} < \frac{1.27 -1.3}{0.01} ) = P(Z < -3) = 1 - P(Z < 3) = 1 - 0.99865

                                                                                            = 0.00135

 P(1.27 < X < 1.33) = 0.99865 - 0.00135 = 0.9973 .

Therefore, proportion of all bearings that falls in this acceptable range is 99.73% .

7 0
3 years ago
A store owner puts in an order for product X and product Y for a total of 4,000 units. It costs $0.10 per item to ship of produc
lyudmila [28]

Answer:

0.1x+0.04y=4,000

x+y=352

Step-by-step explanation:

5 0
3 years ago
Anderson Systems is considering a project that has the following cash flow and WACC data. What is the project's NPV? Note that i
AURORKA [14]

Answer:

$265.65

Step-by-step explanation:

Given :

WACC: 9.00%

Year 0 1 2 3 Cash flows (-$1,000) $500 $500 $500

The NPV is calculated thus :

Initial cashflow + Σ additional cash flows / (1 + r)

Rate, r = 9% = 0.09

(1 + r) = (1 + 0.09) = 1.09

NPV = - 1000 + (500 / (1.09)¹ + (500 / 1.09)² + (500 / (1.09)³

NPV = - 1000 + 458.71559 + 420.83999 + 386.09174

NPV = 265.64732

NPV = 265.65 (2 DECIMAL PLACES)

5 0
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Pavlova-9 [17]
A. 2(-2) - 3
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-7

b. 2(7) - 3
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11

2(-4) - 3
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5 0
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