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3 years ago

What is the point slope form of Y=1/2X-2

1 answer:

8 0

Point-slope form of a line: We need a point (x₀,y₀) and the slope "m";
y-y₀=m(x-x₀)

We have the next equation of line:
y=1/2 x-2 (slope-intercept form y=mx+b)

the slope of this line is 1/2 (m=1/2)
And any one point could be:
if x=0; then y=1/2 (0)-2=-2 (0,-2)

Therefore, we already have the point (0-,2) and the slope (m=1/2)
y-y₀=m(x-x₀)
y+2=1/2(x-0)

Answer: the point slope form of y=1/2 x-2; would be:
y+2=1/2(x-0)

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Can the sum of two mixed numbers be equal to 2, explain why or why not

Sergeeva-Olga [200]

No because a mixed number is a whole number with a fraction and the smallest whole number is one so 1 and a fraction pulse and a fraction will equal more than 2

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3 years ago

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-4 = - 17 + x linear equations

Llana [10]

add 17 to both sides

13 = x

<em>hope this helps</em>

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2 years ago

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A zookeeper feeds three bales of hay each to 4 elephants, 6 bales of hay to each of 10 pens of reindeer, 2 bales of hay to each

snow_lady [41]

Answer:

$105

Step-by-step explanation:

3x5=15

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=$105

5 0

3 years ago

X and y are normal random variables with e(x) = 2, v(x) = 5, e(y) = 6, v(y) = 8 and cov(x,y)=2. determine the following: e(3x 2y

andriy [413]

The result for the given normal random variables are as follows;

a. E(3X + 2Y) = 18

b. V(3X + 2Y) = 77

c. P(3X + 2Y < 18) = 0.5

d. P(3X + 2Y < 28) = 0.8729

<h3>What is normal random variables?</h3>

Any normally distributed random variable having mean = 0 and standard deviation = 1 is referred to as a standard normal random variable. The letter Z will always be used to represent it.

Now, according to the question;

The given normal random variables are;

E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8.

Part a.

Consider E(3X + 2Y)

$\begin{aligned}E(3 X+2 Y) &=3 E(X)+2 E(Y) \\&=(3) (2)+(2)(6 )\\&=18\end{aligned}$

Part b.

Consider V(3X + 2Y)

$\begin{aligned}V(3 X+2 Y) &=3^{2} V(X)+2^{2} V(Y) \\&=(9)(5)+(4)(8) \\&=77\end{aligned}$

Part c.

Consider P(3X + 2Y < 18)

A normal random variable is also linear combination of two independent normal random variables.

$3 X+2 Y \sim N(18,77)$

Thus,

$P(3 X+2 Y < 18)=0.5$

Part d.

Consider P(3X + 2Y < 28)

$Z=\frac{(3 X+2 Y-18)}{\sqrt{77}}$

$\begin{aligned} P(3X + 2Y < 28)&=P\left(\frac{3 X+2 Y-18}{\sqrt{77}} < \frac{28-18}{\sqrt{77}}\right) \\&=P(Z < 1.14) \\&=0.8729\end{aligned}$

Therefore, the values for the given normal random variables are found.

To know more about the normal random variables, here

brainly.com/question/23836881

#SPJ4

The correct question is-

X and Y are independent, normal random variables with E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8. Determine the following:

a. E(3X + 2Y)

b. V(3X + 2Y)

c. P(3X + 2Y < 18)

d. P(3X + 2Y < 28)

8 0

1 year ago

Jamel watched his older brother Robert lift weights. The bar alone had a mass of 20 kg. On the bar he had tow 11.4 kg weights, t

slavikrds [6]

All you do is add them all together...
20 + 11.4 + 11.4 + 4.5 + 4.5 + 0.45 + 0.45 + 0.45 + 0.45 = 53.6Kg
(note that 0.45 is 450g in Kg, When adding things such as weight or length you want to make sure that they are all in the same unit)

6 0

3 years ago