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garri49 [273]
3 years ago
5

Is this proportional or nonpoportional?​

Mathematics
1 answer:
Romashka-Z-Leto [24]3 years ago
6 0
No I don’t think so (&/&/ need extra characters)
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Express M=cos3A + cos A and N = sin3A + sin A as a product form​
elena-14-01-66 [18.8K]

Answer:

2cosAcos2A, 4sinAcos^2A

Step-by-step explanation:

cos3A+cosA

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What is the value of x in the proportion below?<br><br> 57=x21
Rama09 [41]

Answer:

2.71428571429

Step-by-step explanation:

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3 years ago
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 <em>Y</em> = the time taken to deliver a pizza.

The random variable <em>Y</em> 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 <em>X</em> = results of a certain blood test.

It is provided that the random variable <em>X</em> 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 <em>z</em>-scores.

The <em>z</em>-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%.

4 0
3 years ago
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