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3 years ago

12

I need help with this question like the answer please help

2 answers:

aalyn [17]3 years ago

5 0

3 choice because it said several we don’t actually know it so it’s x

Yuri [45]3 years ago

3 0

It started at 1.85, it grew x, the result was 5.30. That's

Answer: 1.85 + x = 5.30 third choice

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What is the equation of the line that passes through (-9,12) and is perpendicular to the line whose equation is y=1/3x 6?

tigry1 [53]

Y = 1/3x ? 6....slope here is 1/3. A perpendicular line will have a negative reciprocal slope. All that means is " flip " the slope and change the sign. So our perpendicular line will have a slope of -3 (see how I flipped 1/3 and changed the sign)

y = mx + b
slope(m) = -3
(-9,12)...x = -9 and y = 12
now we sub and find b, the y int
12 = -3(-9) + b
12 = 27 + b
12 - 27 = b
-15 = b

so ur perpendicular equation is : y = -3x - 15

6 0

3 years ago

8. Find the values of angles x, y,

Eddi Din [679]

Answer:

option A

Step-by-step explanation:

option A is the correct answer of this question.

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8 0

3 years ago

Read 2 more answers

Please help! What is -9/5 - (-9/5)? A. 0 B.2 C. 18/5 D. -18/5

Pie

Answer:6

Step-by-step explanation:

8 0

4 years ago

Read 2 more answers

An individual who has automobile insurance from a certain company is randomly selected. Let Y be the number of mov- ing violatio

olga2289 [7]

Answer:

a) $E(Y) = \sum_{i=1}^n Y_i P(Y_i)$

And replacing we got:

$E(Y) =0*0.60 +1*0.25 +2*0.1 +3*0.05 = 0.60$

b) $E(Y^2) =0^2*0.60 +1^2*0.25 +2^2*0.1 +3^2*0.05 = 1.1$

And then the expected value would be:

$E(100Y^2) = 100*1.1= 110$

Step-by-step explanation:

We assume the following distribution given:

Y 0 1 2 3

P(Y) 0.60 0.25 0.10 0.05

Part a

We can find the expected value with this formula:

$E(Y) = \sum_{i=1}^n Y_i P(Y_i)$

And replacing we got:

$E(Y) =0*0.60 +1*0.25 +2*0.1 +3*0.05 = 0.60$

Part b

If we want to find the expected value of $100 Y^2$ we need to find the expected value of Y^2 and we have:

$E(Y^2) = \sum_{i=1}^n Y^2_i P(Y_i)$

And replacing we got:

$E(Y^2) =0^2*0.60 +1^2*0.25 +2^2*0.1 +3^2*0.05 = 1.1$

And then the expected value would be:

$E(100Y^2) = 100*1.1= 110$

4 0

3 years ago

Figure a is a scale image of figure b. what is the value of x? please answer asap!

emmainna [20.7K]

<em>Greetings from Brasil...</em>

According to the statement, one figure is scaled in relation to another, so we can apply similarity to polygons.....

So

BIG/small = BIG/small

or

small/BIG = small/BIG

12.5/10 = X/16

OR

10/12.5 = 16/X

12.5/10 = X/16

10X = 12.5 · 16

X = 200/10

<h3>X = 20</h3>

8 0

4 years ago