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3 years ago

Suppose William and Donald both drive the same car, and have the same

Mathematics

1 answer:

nataly862011 [7]3 years ago

5 0

Answer: William

Step-by-step explanation: This is true because whichever has a better gas mileage pays less, so the bigger number per year has better mileage, so 15,000 would have better which is Donald, and William has 12,000 which is worse, so he would have to pay more.

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2) A government agency produced a study in which they randomly chose 100 students just beginning college to determine what perce

just olya [345]

The answer is B. some of the sample data are missing.

This is because out of the 100 students originally sampled by the government agency, only 88 results were taken.

Additionally, we cannot fully take A. as the answer because a non-response error happens when sampling units are not interviewed maybe because they were unavailable or unwilling. So B. is the safer answer.

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2 years ago

2. picking a number from 1 to 5 and choosing<br> the color red, white, or blue

bekas [8.4K]

Answer:

i would choose a 3 and color red

Step-by-step explanation:

i still don't understand the question tho

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2 years ago

I need help please!!!!

nika2105 [10]

Answer: I think it is 364.80

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2 years ago

Jace went shopping for a new phone. The listed price of the phone was $50, but the

sukhopar [10]

Answer:

1.50

Step-by-step explanation:

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2 years ago

Find the area of the part of the plane 3x 2y z = 6 that lies in the first octant.

gavmur [86]

The area of the part of the plane 3x 2y z = 6 that lies in the first octant is mathematically given as

A=3 √(4) units ^2

<h3>What is the area of the part of the plane 3x 2y z = 6 that lies in the first octant.?</h3>

Generally, the equation for is mathematically given as

The Figure is the x-y plane triangle formed by the shading. The formula for the surface area of a z=f(x, y) surface is as follows:

$A=\iint_{R_{x y}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y(1)$

The partial derivatives of a function are f x and f y.

$\begin{aligned}&Z=f(x)=6-3 x-2 y \\&=\frac{\partial f(x)}{\partial x}=-3 \\&=\frac{\partial f(y)}{\partial y}=-2\end{aligned}$

When these numbers are plugged into equation (1) and the integrals are given bounds, we get:

$&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{(-3)^{2}+(-2)^2+1dxdy} \\\\&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{14} d x d y \\\\&=\sqrt{14} \int_{0}^{2}[y]_{0}^{3-\frac{3}{2} x} d x d y \\\\&=\sqrt{14} \int_{0}^{2}\left[3-\frac{3}{2} x\right] d x \\\\$

$&=\sqrt{14}\left[3 x-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3.2-\frac{3}{2} \cdot \frac{1}{2} \cdot 3^{2}\right] \\\\&=3 \sqrt{14} \text { units }{ }^{2}$

In conclusion, the area is

A=3 √4 units ^2