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3 years ago

Write an equation in slope intercept form y-intercept 4 and slope is -3/5

1 answer:

6 0

$\bf y=-\cfrac{3}{5}x+4\qquad \impliedby \qquad \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

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After Mike made deposits of $32.88 and $422.73 and wrote checks for $350.00, $126.13, and $41.09, he had $249.81 in his checking

irina1246 [14]

Answer:

$311.42

Step-by-step explanation:

assuming he would have $0.00 dollars left over after the checks. add the checks together then add the deposits together then subtract.

767.03 - 455.61 = 311.42

8 0

3 years ago

Out of 120 high school boys and 80 high school girls, 45 tried out for track. Of those 45, 12 were girls.

zhannawk [14.2K]

<span>So there are 33 boys who tried out for track. And this is 27.5 % of the total boys. And there 15% of the girls who tried out for the track. And in total 22.5 % of the total population tried out for track</span>

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

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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

2 years ago

Does it matter which dimensions you select for the lenght and the width when you find the area of a rectangle? Explain

Levart [38]

It does not. So Long as the length is multipled by the height aka Long edge multiplied with the short edge, you will obtain the area of the rectangle.

3 0

4 years ago

The Factors shot 5 rolls of film with 36 exposures on each roll. It cost $14.85 to process each roll. How much did it cost for e

kondor19780726 [428]

Your answer is .41, hope this helps

6 0

3 years ago

Read 2 more answers