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2 years ago

Urgent help needed will give brainiest

Mathematics

1 answer:

Brut [27]2 years ago

6 0

We can write the integration domain as

$D = \left\{(x,y) \mid -1 \le y \le 1 \text{ and } 2y-2 \le x \le -y+1\right\}$

so that the integral is

$\displaystyle \iint_D -\sin(y+x) \, dA = \int_{-1}^1 \int_{2y-2}^{-y+1} -\sin(y+x) \, dx \, dy$

Compute the integral with respect to $x$ .

$\displaystyle \int_{2y-2}^{-y+1} -\sin(y+x) \, dx = \cos(y+x)\bigg|_{x=2y-2}^{x=-y+1} \\\\ ~~~~~~~~ = \cos(y+(2y-2)) - \cos(y+(-y+1)) \\\\ ~~~~~~~~ = \cos(3y-2) - \cos(1)$

Compute the remaining integral.

$\displaystyle \int_{-1}^1 (\cos(3y-2) - \cos(1)) \, dy = \left(\frac13 \sin(3y-2) - \cos(1) y\right) \bigg|_{y=-1}^{y=1} \\\\ ~~~~~~~~ = \left(\frac13 \sin(3-2) - \cos(1)\right) - \left(\frac13 \sin(-3-2) + \cos(1)\right) \\\\ ~~~~~~~~ = \boxed{\frac13 \sin(1) - 2 \cos(1) + \frac13 \sin(5)}$

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34% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a

finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

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

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

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

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

And $\pi$ is the probability of X happening.

In this problem, we have that:

10 student are sampled, so $n = 10$

34% of college students say they use credit cards because of the rewards program, so $\pi = 0.34$

(a) exactly two

This is P(X = 2).

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

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than two

This is $P(X > 2)$ .

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

$P(X \leq 2) + P(X > 2) = 1$

$P(X > 2) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157$

$P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162$

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573$

$P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320$

$P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434$

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82$

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

6 0

3 years ago

determine if the ordered pair is a solution of the equation is (2,4) a solution of y = 10 -3×? true or false

anzhelika [568]

Answer:

False

Step-by-step explanation:

4 = 10 - 3²

4 = 10 - 9

4 = 1

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3 years ago

What is the probability that a data value in a normal distribution is between a z-score of -1.32 and a z-score of -0.3? round yo

elena55 [62]

Answer:

Step-by-step explanation:

Use the table of Z-score, considering it represnts the probability of data to the left of Z-value, the needed probability would be the subtraction of Z -0.3 to Z -1.32 value.

38.209% - 9.342% =29.867%

The neaerest answer is A

4 0

2 years ago

Which of the following statements about the points (-4,6) and (4,-6) are true

gtnhenbr [62]

Answer:

Step-by-step explanation:

3 0

2 years ago

Line segment NY has endpoints N(-11, 5) and Y(3,-3).

777dan777 [17]