State the solution: 6x + 5 - x = 2(x -2) +3x + 7

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

slega [8]

<span>Let r(x,y) = (x, y, 9 - x^2 - y^2)

So, dr/dx x dr/dy = (2x, 2y, 1)

So, integral(S) F * dS
= integral(x in [0,1], y in [0,1]) (xy, y(9 - x^2 - y^2), x(9 - x^2 - y^2)) * (2x, 2y, 1) dy dx
= integral(x in [0,1], y in [0,1]) (2x^2y + 18y^2 - 2x^2y^2 - 2y^4 + 9x - x^3 - xy^2) dy dx
= integral(x in [0,1]) (x^2 + 6 - 2x^2/3 - 2/5 + 9x - x^3 - x/3) dx
= integral(x in [0,1]) (28/5 + x^2/3 + 26x/3 - x^3) dx
= 28/5 + 40/9 - 1/4
= 1763/180 </span>

6 0

3 years ago

The area of the vegetable garden is 0.4 of the community area of 4.5 x 6.2 . What is the area of the vegetable garden?

Gala2k [10]

Answer:

11.16 units²

Step-by-step explanation:

Area of community area=4.5×6.2= 27.9 units²

Area of vegetable garden=27.9×0.4=11.16 units²

5 0

2 years ago

A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.

nika2105 [10]

Answer:

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

$\bar X=2.55$ represent the sample mean for the late time for a flight

$\mu$ population mean

$\sigma=12$ represent the population deviation

n=76 represent the sample size

Confidence interval

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

The confidence interval for the true mean is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The Confidence level given is 0.95 or 95%, th significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ . If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got $z_{\alpha/2}=1.96$

Replacing we got:

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

7 0

3 years ago

Calculate the straight line distance between the points (-10, -7) and (-8,1).<br> G.2(B)

tia_tia [17]

Answer:

2√(17) or about 8.2462 units.

Step-by-step explanation:

We want to determine the distance between the two points (-10, -7) and (-8, 1).

We can use the distance formula. Recall that:

$\displaystyle d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Substitute and evaluate:

$\displaystyle \begin{aligned} d &= \sqrt{((-10)-(-8))^2 + ((-7) - (1))^2} \\ \\ &=\sqrt{(-2)^2+(-8)^2} \\ \\ &= \sqrt{(4) + (64)} \\ \\ &= \sqrt{68}\\ \\ &= 2\sqrt{17}\end{aligned}$

Hence, the distance between (-10, -7) and (-8, 1) is 2√(17) units or about 8.2462 units.

4 0

2 years ago

A swimsuit is in sale for 45.50. if the sales price is discounted 5%from the orginal price. what was the original price. round t

AnnyKZ [126]

The answer is 42.80 , 2.28 is what they saved and i got the .80 cents because i rounded up from .78

8 0

3 years ago