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3 years ago

Unit 0 algebra review functions graphing linear equations

1 answer:

3 0

Linear Equations:

y = mx + b

The y will of course, be the y coordinate

The m, would be the slope

X, the x coordinate

B, the y intercept

I'm not really sure what your question is, but to graph these, first put a dot at the y intercept, or if there is none, just 0. Then, take the slope, and place a point accordingly, for example, if the slope was 1, then the line would move up one unit per every x coordinate, if that makes any sense. And then, connect the two, and draw the line.

Did that make any sense?

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What is the product of 57.2 and 32

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3 years ago

Read 2 more answers

Assume that you want to test the claim that the paired sample data come from a population for which the mean difference is μd =

swat32

Answer:

Test statistic, $t_{s} = -0.603$ (to 3 dp)

Step-by-step explanation:

Deviation, d = x -y

Sample mean for the deviation

$\bar{d} = \frac{\sum x-y}{n}$

$\bar{d} = \frac{(28-6) + (31-27)+(20-26)+(25-25)+(28-29)+(27-32)+(33-33)+(35-34)}{8} \\\bar{d} = -0.625$

Standard deviation: $SD = \sqrt{\frac{\sum d^{2} - n \bar{d}^2}{n-1} }$

$\sum d^{2} = (28-26)^2 + (31-27)^2 +(20-26)^2 +(25-25)^2 +(28-29)^2 +(27-32)^2 +(33-33)^2 +(35-34)^2\\\sum d^{2} = 63$

$SD = \sqrt{\frac{63 - 8 * (-0.625)^2}{8-1} }$

SD =2.93

Under the null hypothesis, the formula for the test statistics will be given by:

$t_{s} = \frac{ \bar{d}}{s_{d}/\sqrt{n} } \\t_{s} = \frac{- 0.625}{2.93/\sqrt{8} }$

$t_{s} = -0.6033$

6 0

3 years ago

A manufacturer is planning to sell a new product at the price of 210 dollars per unit and estimates that if x thousand dollars i

sattari [20]

Answer:

Development costs: $2757

Promotion costs: $3155

Step-by-step explanation:

This might seem like a two-variable problem, but in actuality it's not - because the demand function is a sum of two components, each being independent and using only one variable, we can solve for the two separately.

Moreover, the price per unit and cost per unit is constant, so each product yields exactly $80 ($210 - $130).

Let's solve separately.

We need to maximize:

$80 * 160y/(y+4) - 1000y = $12800 * y/(y+4) - $1000*y

$80 * 170x/(x+7) - 1000x = $13600 * x/(x+7) - $1000*x

Let's go:

$12800 * y/(y+4) - $1000*y = $12800 * (1 - 4/(y+4)) - $1000y = $12800 - $51200/(y+4) - $1000*y

we analyze the derivative. $12800 is a constant, so we can skip it. Derivative of 1/(y+4) is -(y+4)^-2, derivative of $1000y is $1000.

deriv = $51200/(y+4)/(y+4) - $1000

We find the changepoints by analyzing $51200/(y+4)/(y+4) - $1000 = $0. We don't need to worry about y+4 = 0 because we cannot spend negative money on development/advertisement.

(y+4)^2 = $51.2

y+4 ~= 7.155417528 (or -y-4 = 7.1554... but it doesn't make sense because negative budget so we don't analyze).

y ~= 3.155417528

lastly, we should check that it's actually a maximum there - but it is, the original function goes to negative infinity.

rounding $1000y to the nearest dollar gives us $3155

Let's do the same for x:

$13600 * x/(x+7) - $1000*x = $13600 * (1 - 7/(x+7)) - $1000x = $13600 - $95200/(x+7) - $1000*x

deriv = $95200/(x+7)/(x+7) - $1000

$95200/(x+7)/(x+7) - $1000 = $0

(x+7)^2 = 95.2

(x+7) ~= 9.75704873412

x ~= 2.75704...

rounding $1000x to the nearest dollar yields $2757

3 0

3 years ago