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2 years ago

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Which equation represents a line that passes through (–2, 4) and has a slope of StartFraction 2 Over 5 EndFraction.? y – 4 = (x

+ 2) y + 4 =(x – 2) y + 2 = (x – 4) y – 2 =(x +4)

2 answers:

JulijaS [17]2 years ago

8 0

For this case we have that by definition, the point-slope equation of a line is given by:

$(y-y_ {0}) = m (x-x_ {0})$

Where:

$(x_ {0}, y_ {0}):$ It is a point through which the line passes

m: It is the slope of the line

According to the data we have to:

$(x_ {0}, y_ {0}): (-2,4)\\m = \frac {2} {5}$

Substituting:

$y-4 = \frac {2} {5} (x - (- 2))\\y-4 = \frac {2} {5} (x + 2)$

Finally, the equation of the line is:

$y-4 = \frac {2} {5} (x + 2)$

ANswer:

$y-4 = \frac {2} {5} (x + 2)$

levacccp [35]2 years ago

7 0

Answer:

The answer is A

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2 years ago

An certain brand of upright freezer is available in three different rated capacities: 16 ft3, 18 ft3, and 20 ft3. Let X = the ra

ruslelena [56]

Answer:

$E(X)=16\cdot 0.3+18\cdot 0.1+20\cdot 0.6=18.6$

$E(X^2)=16^2\cdot 0.3+18^2\cdot 0.1+20^2\cdot 0.6=349.2$

$V(X) = E(X^2)-[E(X)]^2=349.2-(18.6)^2=3.24$

The expected price paid by the next customer to buy a freezer is $466

Step-by-step explanation:

From the information given we know the probability mass function (pmf) of random variable X.

$\left|\begin{array}{c|ccc}x&16&18&20\\p(x)&0.3&0.1&0.6\end{array}\right|$

<em>Point a:</em>

The Expected value or the mean value of X with set of possible values D, denoted by <em>E(X)</em> or <em>μ </em>is

$E(X) = $\sum_{x\in D} x\cdot p(x)$

Therefore

$E(X)=16\cdot 0.3+18\cdot 0.1+20\cdot 0.6=18.6$

If the random variable X has a set of possible values D and a probability mass function, then the expected value of any function h(X), denoted by <em>E[h(X)]</em> is computed by

$E[h(X)] = $\sum_{D} h(x)\cdot p(x)$

So $h(X) = X^2$ and

$E[h(X)] = $\sum_{D} h(x)\cdot p(x)\\E[X^2]=$\sum_{D}x^2\cdot p(x)\\ E(X^2)=16^2\cdot 0.3+18^2\cdot 0.1+20^2\cdot 0.6\\E(X^2)=349.2$

The variance of X, denoted by V(X), is

$V(X) = $\sum_{D}E[(X-\mu)^2]=E(X^2)-[E(X)]^2$

Therefore

$V(X) = E(X^2)-[E(X)]^2\\V(X)=349.2-(18.6)^2\\V(X)=3.24$

<em>Point b:</em>

We know that the price of a freezer having capacity X is 60X − 650, to find the expected price paid by the next customer to buy a freezer you need to:

From the rules of expected value this proposition is true:

$E(aX+b)=a\cdot E(x)+b$

We have a = 60, b = -650, and <em>E(X)</em> = 18.6. Therefore

The expected price paid by the next customer is

$60\cdot E(X)-650=60\cdot 18.6-650=466$

4 0

2 years ago

Kaitlyn solved the equation for x using the following calculations. Negative 5.4 + x = 12.2. Minus 5.4. x = 6.8. What error, if

omeli [17]

Step-by-step explanation:

The equation that Kaitlyn solved is as follows :

-5.4+x=12.2

The steps to solve it as follows :

Step 1. Add 5.4 to both sides,

-5.4+x+5.4=12.2+5.4

Step 2. 5.4 will cancel out as both have opposite signs

x=17.6

But Kaitlyn solved it as :

5.4.x=6.8

She has multiplied -5.4 and x which is wrong. The correct answer is 17.6.

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2 years ago

Which graph represents viable values for y = 2x, where x is the number of pounds of rice scooped and purchased from a bulk bin a

AleksAgata [21]

Answer:

On a coordinate plane, a straight line with a positive slope begins at point (0, 0), and ends at point (2.5, 5)

Step-by-step explanation:

The distinction between "straight line" and "blue diamonds" is that the straight line represents the relation for all possible values of x. The blue diamonds show the values of y for very specific values of x.

Since x is the amount of rice from a bulk bin, we assume it can take any non-negative value. Hence, the graphs with negative values or with "blue diamonds" are not appropriate.

The straight-line graph in the first quadrant is the best choice.

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2 years ago

Read 2 more answers

World poultry production was 77.2 million tons in the year 2004 and increasing at a continuous rate of 1.6% per year. Assume tha

dezoksy [38]

Answer:

A) $P(t)=77.2\cdot e^{0.016t}$

B) 89.157 million tons.

C) In year 2017.

Step-by-step explanation:

We have been given that world poultry production was 77.2 million tons in the year 2004 and increasing at a continuous rate of 1.6% per year.

A). We know that a continuous growth function is in form $y=a\cdot e^{kx}$ , where,

a = Initial value,

k = Growth rate in decimal form.

$1.6\%=\frac{1.6}{100}=0.016$

Upon substituting our given values, we will get:

$P(t)=77.2\cdot e^{0.016t}$

Therefore, our required function would be $P(t)=77.2\cdot e^{0.016t}$ , where, P represents world poultry production, in million tons, as a function of the number of years, t, since 2004.

B) To find the world poultry production in the year 2013, we will substitute $t=2013-2004=9$ in above formula as:

$P(9)=77.2\cdot e^{0.016(9)}$

$P(9)=77.2\cdot e^{0.144}$

$P(9)=77.2\cdot 1.1548841085249135$

$P(9)=89.1570531781\approx 89.157$

Therefore, the world poultry production in the year 2013 would be 89.157 million tons.

C) To find the number of years it will take for world poultry production to be over 95 million tons, we will equate our function with 95 as:

$95=77.2\cdot e^{0.016t}$

Divide both sides by 77.2:

$1.2305699481865285=e^{0.016t}$

Take natural log of both sides:

$\text{ln}(1.2305699481865285)=\text{ln}(e^{0.016t})$

$\text{ln}(1.2305699481865285)=0.016t\text{ln}(e)$

$0.20747743456981035=0.016t*1$

Divide both sides by 0.016:

$t=12.9673396606$

$t\approx 13$

Therefore, 13 years after in 2017 world poultry production goes over 95 million tons.

4 0

2 years ago