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Akimi4 [234]
3 years ago
9

The graph of line p represents y =1/5 x-1 . If the slope of line p is multiplied by -10 to

Mathematics
1 answer:
Katarina [22]3 years ago
5 0

Answer:

use desmos its a graphing caculator that gives you exaxt graphing answers

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Caroline has 16 marbles 1\8 of them are blue how many of Caroline's marbles are blue
seropon [69]
2 of them is blue because since there is 16 marbles you multiply 1/8 by 2 and 2/16 so 2 marbles are blue 
7 0
3 years ago
Read 2 more answers
What is the x-intercept (-5,-5) (10,-2)
Andre45 [30]

Answer:

(20,0)

Step-by-step explanation:

slope = (y2 - y1) / (x2 - x1)

(-5,-5).....x1 = -5 and y1 = -5

(10,-2)...x2 = 10 and y2 = -2

slope = (-2 - (-5) / (10- (-5) = (-2 + 5) / (10 + 5) = 3/15 = 1/5

y = mx + b

slope(m) = 1/5

use either point, I will use (10,-2)...x = 10 and y = -2

now we sub

-2 = 1/5(10) + b

-2 = 2 + b

-2 - 2 = b

-4 = b

so the equation for this line is : y = 1/5x - 4

to find the x intercept, sub in 0 for y and solve for x

y = 1/5x - 4

0 = 1/5x - 4

-1/5x = -4

x = -4 * -5

x = 20.......so ur x intercept is (20,0) <====

6 0
3 years ago
I need the answer with work
MakcuM [25]
The answer to what? If you can tell me, I'd be happy to help. :)
3 0
3 years ago
A city planner wants to build a road parallel to 2nd Ave . What is the slope of the new road?
ludmilkaskok [199]

Answer:

0

Step-by-step explanation:

By inspection or finding points, we see that 2nd ave is perfectly flat, or in other words, has slope of 0.

If line 1 is parallel to line 2, it means their slopes are the same.

So, the new road parallel to 2nd ave has a slope of \boxed{0}.

6 0
3 years ago
A broker has calculated the expected values of two different financial instruments X and Y. Suppose that E(x)= $100, E(y)=$90 SD
Sveta_85 [38]

Expectation is linear, meaning

E(<em>a X</em> + <em>b Y</em>) = E(<em>a X</em>) + E(<em>b Y</em>)

= <em>a </em>E(<em>X</em>) + <em>b</em> E(<em>Y</em>)

If <em>X</em> = 1 and <em>Y</em> = 0, we see that the expectation of a constant, E(<em>a</em>), is equal to the constant, <em>a</em>.

Use this property to compute the expectations:

E(<em>X</em> + 10) = E(<em>X</em>) + E(10) = $110

E(5<em>Y</em>) = 5 E(<em>Y</em>) = $450

E(<em>X</em> + <em>Y</em>) = E(<em>X</em>) + E(<em>Y</em>) = $190

Variance has a similar property:

V(<em>a X</em> + <em>b Y</em>) = V(<em>a X</em>) + V(<em>b Y</em>) + Cov(<em>X</em>, <em>Y</em>)

= <em>a</em>^2<em> </em>V(<em>X</em>) + <em>b</em>^2 V(<em>Y</em>) + Cov(<em>X</em>, <em>Y</em>)

where "Cov" denotes covariance, defined by

E[(<em>X</em> - E(<em>X</em>))(<em>Y</em> - E(<em>Y</em>))] = E(<em>X Y</em>) - E(<em>X</em>) E(<em>Y</em>)

Without knowing the expectation of <em>X Y</em>, we can't determine the covariance and thus variance of the expression <em>a X</em> + <em>b Y</em>.

However, if <em>X</em> and <em>Y</em> are independent, then E(<em>X Y</em>) = E(<em>X</em>) E(<em>Y</em>), which makes the covariance vanish, so that

V(<em>a X</em> + <em>b Y</em>) = <em>a</em>^2<em> </em>V(<em>X</em>) + <em>b</em>^2 V(<em>Y</em>)

and this is the assumption we have to make to find the standard deviations (which is the square root of the variance).

Also, variance is defined as

V(<em>X</em>) = E[(<em>X</em> - E(<em>X</em>))^2] = E(<em>X</em>^2) - E(<em>X</em>)^2

and it follows from this that, if <em>X</em> is a constant, say <em>a</em>, then

V(<em>a</em>) = E(<em>a</em>^2) - E(<em>a</em>)^2 = <em>a</em>^2 - <em>a</em>^2 = 0

Use this property, and the assumption of independence, to compute the variances, and hence the standard deviations:

V(<em>X</em> + 10) = V(<em>X</em>)  ==>  SD(<em>X</em> + 10) = SD(<em>X</em>) = $90

V(5<em>Y</em>) = 5^2 V(<em>Y</em>) = 25 V(<em>Y</em>)  ==>  SD(5<em>Y</em>) = 5 SD(<em>Y</em>) = $40

V(<em>X</em> + <em>Y</em>) = V(<em>X</em>) + V(<em>Y</em>)  ==>  SD(<em>X</em> + <em>Y</em>) = √[SD(<em>X</em>)^2 + SD(<em>Y</em>)^2] = √8164 ≈ $90.35

8 0
4 years ago
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