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3 years ago

8

.can somebody help me.with this question

1 answer:

oksian1 [2.3K]3 years ago

8 0

Answer:

4 and 16

Step-by-step explanation:

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If E(X)=100, E(Y)=120, E(Z) = 130, Var(X) = 9, Var(Y) = 16, Var(Z) = 25, Cov(X, Y)= - 10 Cov(X,Z) = 12, and Cov(Y,Z) = 14, then

vredina [299]

Answer:

(1) -0.833

(2) 0.80

(3) 0.70

(4) 390

(5) 90

(7) 48

Step-by-step explanation:

Given:

E (X) = 100, E (Y) = 120, E (Z) = 130

Var (X) = 9, Var (Y) = 16, Var (Z) = 25

Cov (X, Y) = -10, Cov (X, Z) = 12, Cov (Y, Z) = 14

The formulas used for correlation is:

$Corr (A, B) = \frac{Cov (A, B)}{\sqrt{Var (A)\times Var(B)}} \\$

(1)

Compute the value of Corr (X, Y)-

$Corr (X, Y) = \frac{Cov (X, Y)}{\sqrt{Var (X)\times Var(Y)}} \\=\frac{-10}{\sqrt{9\times16}} \\=-0.833$

(2)

Compute the value of Corr (X, Z)-

$Corr (X, Z) = \frac{Cov (X, Z)}{\sqrt{Var (X)\times Var(Z)}} \\=\frac{12}{\sqrt{9\times25}} \\=0.80$

(3)

Compute the value of Corr (Y, Z)-

$Corr (Y, Z) = \frac{Cov (Y, Z)}{\sqrt{Var (Y)\times Var(Z)}} \\=\frac{14}{\sqrt{16\times25}} \\=0.70$

(4)

Compute the value of E (3X+4Y-3Z)-

$E(3X+4Y-3Z)=3E(X)+4E(Y)-3E(Z)\\=(3\times100)+(4\times120)-(3\times130)\\=390$

(5)

Compute the value of Var (3X-3Z)-

$Var (3X-3Z)=[(3)^{2}\times Var(X)]+[(-3)^{2}\times Var (Z)]+(2\times3\times-3\times Cov(X, Z)]\\=(9\times9)+(9\times25)-(18\times12)\\=90$

(6)

Compute the value of Var (3X+4Y-3Z)-

$Var (3X+4Y-3Z)=[(3)^{2}\times Var(X)]+[(4)^{2}\times Var(Y)]+[(-3)^{2}\times Var (Z)]+[(2\times3\times4\times Cov(X, Y)]+[(2\times3\times-3\times Cov(X, Z)]+[(2\times4\times-3\times Cov(Y, Z)]\\=(9\times9)+(16\times16)+(9\times25)+(24\times-10)-(18\times12)-(24\times14)\\=-230$

But this is not possible as variance is a square of terms.

(7)

Compute the value of Cov (3X, 2Y+3Z)-

$Cov(3X, 2Y+3Z)=Cov(3X,2Y)+Cov(3X, 3Z)\\=6Cov(X, Y)+9Cov(X,Z)\\=(6\times-10)+(9\times12)\\=48$

4 0

3 years ago

Find the equation of the straight line passing through the points (-1,3) and (1,9).

Lena [83]

Answer:

The equation of the straight line is $y=3x+6$

Step-by-step explanation:

Any straight line passing through points $(a_1,b_1)$ and $(a_2,b_2)$ is given by:

$y-b_1=m(x-a_1)$ where m is defined as the slope of the line:

$m=\frac{b_2-b_1}{a_2-a_1}$

In the given question

$m=\frac{9-3}{1-(-1)} =3$

Substituting the values we get:

<h2><u></u> $y-3=3(x+1)\\y-3=3x+3\\y=3x+6$ </h2>

Reference: https://www.cuemath.com/geometry/equation-of-a-straight-line/

#SPJ2

6 0

2 years ago

(6y^2-9y+4)-(-7y^2+5y+1)

Vladimir79 [104]

Answer and explanation:

$(6y^2-9y+4)-(-7y^2+5y+1)$ multiply the - into $-7y^2+5y+1$

$6y^2-9y+4+7y^2-5y-1$ Now add, or subtract, everything together.

$(13y^2-14y+3)$ Answer

3 0

4 years ago

At a local department store, sweaters are typically priced at $25. Due to a special, the sweaters are reduced to 80% of their or

WITCHER [35]

Multiply 25 by .80 and subtract that number from $25.
25x0.80= 20. 25-20= $5
Hope I helped!

4 0

3 years ago

A sedan will drive ten miles north, 15 miles east, 13 miles south, 15 miles west, and 22 miles north. The sedan gets 20 miles pe

jarptica [38.1K]

<u>Answer</u>:

There is enough fuel for the trip and the relative distance is 19 miles

<u>Explanation</u>:

Given Sedan drives

10 miles north

15 miles east

13 miles south

15 miles west

22 miles north

Then total distance = 10 + 15 + 13 + 15 + 22 = 75 miles

There are 5 gallons of gas left and sedan gets 20 miles per gallon

20 miles = 1 gallon

75 miles = 75/20 = 3.75 gallons

Gas left = 5 – 3.75 = 1.25 gallon

So, there is enough fuel for the trip

Relative distance = 10 (north) + 15 (east) – 13 (south) – 15 (west) + 22 (north) = 19 miles

7 0

3 years ago