A TV station has 11 minutes of commercials for each 60 minutes of air time. What percent

$D$ is a right triangle with base length 1 and height 8, so the area of $D$ is $\dfrac12(1)(8)=4$ .

The average value of $f(x,y)$ over $D$ is given by the ratio

$\dfrac{\displaystyle\iint_Df(x,y)\,\mathrm dA}{\displaystyle\iint_D\mathrm dA}$

The denominator is just the area of $D$ , which we already know. The average value is then simplified to

$\displaystyle\frac74\iint_Dxy\,\mathrm dA$

In the $x,y$ -plane, we can describe the region $D$ as all points $(x,y)$ that lie between the lines $y=0$ and $y=8x$ (the lines which coincide with the triangle's base and hypotenuse, respectively), taking $0\le x\le1$ . So, the integral is given by, and evaluates to,

$\displaystyle\frac74\int_{x=0}^{x=1}\int_{y=0}^{y=8x}xy\,\mathrm dy\,\mathrm dx=\frac78\int_{x=0}^{x=1}xy^2\bigg|_{y=0}^{y=8x}\,\mathrm dx$
$=\displaystyle56\int_{x=0}^{x=1}x^3\,\mathrm dx$
$=14x^4\bigg|_{x=0}^{x=1}$
$=14$

3 0

3 years ago

Wday is the area in square inches of the floor in the scale drawing

Mice21 [21]

Answer:

8

Step-by-step explanation:

Since 24 is the original length, you would have to see what number to divide by to get 6.

24 divided by 4 is 6.

Now we know that 1/4 is the scale factor, so apply that to 32.

32 divided by 4 is 6

That's your answer!

Brainliest would v much be appreciated

5 0

3 years ago

12. Phillip deposited $450 in an account that earns

Salsk061 [2.6K]

This is your genuine answer

7 0

3 years ago

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(10 points)Assume IQs of adults in a certain country are normally distributed with mean 100 and SD 15. Suppose a president, vice

vesna_86 [32]

Answer:

0.0139 = 1.39% probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

Step-by-step explanation:

To solve this question, we need to use the binomial and the normal probability distributions.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Probability the president will have an IQ of at least 107.5

IQs of adults in a certain country are normally distributed with mean 100 and SD 15, which means that $\mu = 100, \sigma = 15$

This probability is 1 subtracted by the p-value of Z when X = 107.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{107.5 - 100}{15}$

$Z = 0.5$

$Z = 0.5$ has a p-value of 0.6915.

1 - 0.6915 = 0.3085

0.3085 probability that the president will have an IQ of at least 107.5.

Probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

First, we find the probability of a single person having an IQ of at least 130, which is 1 subtracted by the p-value of Z when X = 130. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{130 - 100}{15}$

$Z = 2$

$Z = 2$ has a p-value of 0.9772.

1 - 0.9772 = 0.0228.

Now, we find the probability of at least one person, from a set of 2, having an IQ of at least 130, which is found using the binomial distribution, with p = 0.0228 and n = 2, and we want:

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.9772)^{2}.(0.0228)^{0} = 0.9549$

$P(X \geq 1) = 1 - P(X = 0) = 0.0451$

0.0451 probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

What is the probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130?

0.3085 probability that the president will have an IQ of at least 107.5.

0.0451 probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

Independent events, so we multiply the probabilities.

0.3082*0.0451 = 0.0139

0.0139 = 1.39% probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

8 0

2 years ago

I need help with this question please?!!!

scoundrel [369]

The answer is -1 because it is subtracting by -2

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3 years ago

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