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Ilya [14]
2 years ago
6

What is the system of inequalities associated with the following graph?

Mathematics
1 answer:
GarryVolchara [31]2 years ago
3 0

Answer: C

Step-by-step explanation:

y=-2 is dotted and shaded below, so it represents y < -2.

Also, x+y=1 is shaded above, so it represents x+y < 1.

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Terri Vogel, an amateur motorcycle racer, averages 129.71 seconds per 2.5 mile lap (in a 7 lap race) with a standard deviation o
Vikki [24]

Answer:

(a) The percent of her laps that are completed in less than 130 seconds is 55%.

(b) The fastest 3% of her laps are under 125.42 seconds.

(c) The middle 80% of her laps are from <u>126.80</u> seconds to <u>132.63</u> seconds.

Step-by-step explanation:

The random variable <em>X</em> is defined as the number of seconds for a randomly selected lap.

The random variable <em>X </em>is normally distributed with mean, <em>μ</em> = 129.71 seconds and standard deviation, <em>σ</em> = 2.28 seconds.

Thus, X\sim N(129.71,\ 2.28^{2}).

(a)

Compute the probability that a lap is completes in less than 130 seconds as follows:

P(X

                   =P(Z

The percentage is, 0.55 × 100 = 55%.

Thus, the percent of her laps that are completed in less than 130 seconds is 55%.

(b)

Let <em>x</em> represents the 3rd percentile.

That is, P (X < x) = 0.03.

⇒ P (Z < z) = 0.03

The value of <em>z</em> for the above probability is:

<em>z</em> = -1.88

Compute the value of <em>x</em> as follows:

z=\frac{x-\mu}{\sigma}\\-1.88=\frac{x-129.71}{2.28}\\x=129.71-(1.88\times 2.28)\\x=125.4236\\x\approx 125.42

Thus, the fastest 3% of her laps are under 125.42 seconds.

(c)

Let <em>x</em>₁ and <em>x</em>₂ be the values between which the middle 80% of the distribution lie.

That is,

P(x_{1}

The value of <em>z</em> for the above probability is:

<em>z</em> = 1.28

Compute the values of <em>x</em>₁ and <em>x</em>₂ as follows:

-z=\frac{x_{1}-\mu}{\sigma}\\-1.28=\frac{x_{1}-129.71}{2.28}\\x_{1}=129.71-(1.28\times 2.28)\\x=126.7916\\x\approx 126.80               z=\frac{x_{2}-\mu}{\sigma}\\1.28=\frac{x_{2}-129.71}{2.28}\\x_{2}=129.71+(1.28\times 2.28)\\x=132.6284\\x\approx 132.63

Thus, the middle 80% of her laps are from <u>126.80</u> seconds to <u>132.63</u> seconds.

5 0
3 years ago
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