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3 years ago

If f(x) = x^2 — 3x, explain how to find f(—1.8).

Mathematics

1 answer:

gogolik [260]3 years ago

5 0

Answer:

Plug -1.8 into each x so the equation would become (-1.8)^2 - 3(-1.8)

Step-by-step explanation:

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What are the coordinates of the point that is common to the graph of f(x) = (2/3) ^ x and the graph of f(x) = (3/2) ^ x

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2 years ago

Maya has 5 over 6 cup of ice cream. How many 1 over 4 -cup servings are in 5 over 6 cup of ice cream?

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Jada has 7/8 cup of cheese. Her cheese bread recipe calls for 1/6 cup of cheese . How many times can she make her recipe with th

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3 years ago

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The time taken to deliver a pizza has a uniform probability distribution from 20 minutes to 60 minutes. What is the probability

Whitepunk [10]

Answer:

(1) The probability that the time to deliver a pizza is at least 32 minutes is 0.70.

(2a) The percentage of results more than 45 is 79.67%.

(2b) The percentage of results less than 85 is 91.77%.

(2c) The percentage of results are between 75 and 90 is 15.58%.

(2d) The percentage of results outside the healthy range 20 to 100 is 2.64%.

Step-by-step explanation:

(1)

Let Y = the time taken to deliver a pizza.

The random variable Y follows a Uniform distribution, U (20, 60).

The probability distribution function of a Uniform distribution is:

$f(x)=\left \{ {{\frac{1}{b-a};\ x\in [a, b] } \atop {0};\ otherwise} \right.$

Compute the probability that the time to deliver a pizza is at least 32 minutes as follows:

$P(Y\geq 32)=\int\limits^{60}_{32} {\frac{1}{b-a} } \, dx \\=\frac{1}{60-20} \int\limits^{60}_{32} {1 } \, dx\\=\frac{1}{40}\times[x]^{60}_{32}\\=\frac{1}{40}\times[60-32]\\=0.70$

Thus, the probability that the time to deliver a pizza is at least 32 minutes is 0.70.

(2)

Let X = results of a certain blood test.

It is provided that the random variable X follows a Normal distribution with parameters $\mu = 60$ and $s = 18$ .

The probabilities of a Normal distribution are computed by converting the raw scores to z-scores.

The z-scores follows a Standard normal distribution, N (0, 1).

(a)

Compute the probability that the results are more than 45 as follows:

$P(X>45)=P(\frac{X-\mu}{\sigma}> \frac{45-60}{18})=P(Z>-0.833)=P(Z$

The percentage of results more than 45 is: $0.7967\times100=79.67\%$

Thus, the percentage of results more than 45 is 79.67%.

(b)

Compute the probability that the results are less than 85 as follows:

$P(X$

The percentage of results less than 85 is: $0.9177\times100=91.77\%$

Thus, the percentage of results less than 85 is 91.77%.

(c)

Compute the probability that the results are between 75 and 90 as follows:

$P(75$

The percentage of results are between 75 and 90 is: $0.1558\times100=15.58\%$

Thus, the percentage of results are between 75 and 90 is 15.58%.

(d)

Compute the probability that the results are between 20 and 100 as follows:

$P(20$

Then the probability that the results outside the range 20 to 100 is: $1-0.9736=0.0264$ .

The percentage of results outside the range 20 to 100 is: $0.0264\times100=2.64\%$

Thus, the percentage of results outside the healthy range 20 to 100 is 2.64%.

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3 years ago