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Ahat [919]
2 years ago
15

Estimate the volume of the solid that lies below the surface z = xy and above the following rectangle. R = (x, y) | 2 ≤ x ≤ 8, 6

≤ y ≤ 10 (a) Use a Riemann sum with m = 3, n = 2, and take the sample point to be the upper right corner of each square. (b) Use the Midpoint Rule to estimate the volume of the solid.

Mathematics
1 answer:
Marianna [84]2 years ago
8 0

See attached picture:

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NEED HELP ASAP!!! ON A KHAN ACADEMY TEST AND DON"T HAVE MUCH TIME LEFT!!! CORRECT ANSWER GETS BRAINLIEST!!!
Otrada [13]

Answer:

Step-by-step explanation:

This is correct, except that if you have two variables you will need two equations. The other equation (the easy one!) was x+ y = 100, from which you might want to solve for y = 100-x, and substitute into the other equation to get:

.30x + .02(100-x) = .04(100)

.30x + 2 - .02x = 4

 

.28x = 2

x = 2/.28 = 50/7 or approximately 7.14 gallons of cream

100-x = 92 6/7 or approximately 92.86 gallons of 2%

R^2 at SCC

6 0
2 years ago
g An irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the
blagie [28]

Answer:

A 90% confidence interval of the true mean is [$119.86, $123.34].

Step-by-step explanation:

We are given that an irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the mean amount of money spent on textbooks was $121.60.

Also, the standard deviation of the population was $6.36.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                              P.Q.  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean amount of money spent on textbooks = $121.60

            \sigma = population standard deviation = $6.36

            n = sample of students = 36

            \mu = population mean

<em>Here for constructing a 90% confidence interval we have used One-sample z-test statistics as we know about population standard deviation.</em>

<em />

So, 95% confidence interval for the population mean, \mu is ;

P(-1.645 < N(0,1) < 1.645) = 0.90  {As the critical value of z at 5% level

                                                      of significance are -1.645 & 1.645}  

P(-1.645 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.645) = 0.90

P( -1.645 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.645 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.90

P( \bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.90

<u>90% confidence interval for</u> \mu = [ \bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } } ]

                                      = [ 121.60-1.645 \times {\frac{6.36}{\sqrt{36} } } , 121.60+1.645 \times {\frac{6.36}{\sqrt{36} } } ]

                                      = [$119.86, $123.34]

Therefore, a 90% confidence interval of the true mean is [$119.86, $123.34].

5 0
2 years ago
Jan accidentally ran 7 minutes longer than she was supposed to. Write an expression for the total amount of time Jan ran if she
vovangra [49]

Answer:

m+7

Step-by-step explanation:

If she ran 7 minutes longer and m represents the amount of time she ran, then it would just be addition.

Let's say m=60 minutes. She would've ran for 67 minutes, which is also m+7.

6 0
2 years ago
Harriet would like to frame her graduation picture. The picture is 8 inches wide and 11 inches long. She wants the total area of
LekaFEV [45]

Answer: i explained

Step-by-step explanation: Just do

8 times 11 = *your answer*

then

*your answer* times/subtract/divided by  108

8 0
2 years ago
3) George walks away from his house. He walks 25 km west and then
Romashka-Z-Leto [24]

Answer:

29.15 km

Step-by-step explanation:

Given;

George walks; 25km west and then 15 km south

Resolving the directions to x and y axis;

North and South represent positive and negative y axis.

East and West represent positive and negative x axis respectively.

25km west

Rx = -25 km

15 km south

Ry = -15 km

The resultant displacement from the house is;

R = √(Rx^2 + Ry^2)

Substituting the values;

R = √((-15)^2 + (-25)^2)

R = √(225+625)

R = √(850)

R = 29.15 km

Therefore, he is 29.15 km from house

8 0
2 years ago
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