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3 years ago

Help please . Thank you

Mathematics

1 answer:

MArishka [77]3 years ago

7 0

The general form is ax²+bx+c=0,
so at least we know b has to be 0. None of the values match unless we multiply everything by 4. That leads to answer B, a=1, b=0 and c=20

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Let Y1 and Y2 have the joint probability density function given by:

Ann [662]

Answer:

a) k=6

b) P(Y1 ≤ 3/4, Y2 ≥ 1/2) = 9/16

Step-by-step explanation:

a) if

f (y1, y2) = k(1 − y2), 0 ≤ y1 ≤ y2 ≤ 1, 0, elsewhere

for f to be a probability density function , has to comply with the requirement that the sum of the probability of all the posible states is 1 , then

P(all possible values) = ∫∫f (y1, y2) dy1*dy2 = 1

then integrated between

y1 ≤ y2 ≤ 1 and 0 ≤ y1 ≤ 1

∫∫f (y1, y2) dy1*dy2 = ∫∫k(1 − y2) dy1*dy2 = k ∫ [(1-1²/2)- (y1-y1²/2)] dy1 = k ∫ (1/2-y1+y1²/2) dy1) = k[ (1/2* 1 - 1²/2 +1/2*1³/3)- (1/2* 0 - 0²/2 +1/2*0³/3)] = k*(1/6)

then

k/6 = 1 → k=6

P(Y1 ≤ 3/4, Y2 ≥ 1/2) = P (0 ≤Y1 ≤ 3/4, 1/2 ≤Y2 ≤ 1) = p

then

p = ∫∫f (y1, y2) dy1*dy2 = 6*∫∫(1 − y2) dy1*dy2 = 6*∫(1 − y2) *dy2 ∫dy1 =

6*[(1-1²/2)-((1/2) - (1/2)²/2)]*[3/4-0] = 6*(1/8)*(3/4)= 9/16

therefore

P(Y1 ≤ 3/4, Y2 ≥ 1/2) = 9/16

8 0

3 years ago

HURRY!!!!<br> What is the equation of the line shown in this graph?

zalisa [80]

Answer:

x = - 2

Step-by-step explanation:

The equation of a vertical line parallel to the y- axis is

x = c

Where c is the value of the x- coordinates the line passes through.

The line passes through (- 2, - 3) and (- 2, 2)

Both of these points have an x- coordinate of - 2, thus

x = - 2 ← equation of line

3 0

3 years ago

Given the function h(x) =1/3 |x-6| +4, evaluate the function when x = - 3, - 2, and 0

Masja [62]

|x| = x for x ≥ 0

examples:

|3| = 3; |0.56| = 0.56; |102| = 102

|x| = -x for x < 0

examples:

|-3| = -(-3) = 3; |-0.56| = -(-0.56) = 0.56; |-102| = 102

--------------------------------------------------------------------------------

Use PEMDAS:

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

--------------------------------------------------------------------------------

$h(x)=\dfrac{1}{3}|x-6|+4$

Put the values of x to the equation of the function h(x):

$x=-3\to h(-3)=\dfrac{1}{3}|-3-6|+4=\dfrac{1}{3}|-9|+4=\dfrac{1}{3}(9)+4=3+4=7\\\\x=-2\to h(-2)=\dfrac{1}{3}|-2-6|+4=\dfrac{1}{3}|-8|+4=\dfrac{1}{3}(8)+4=\dfrac{8}{3}+\dfrac{12}{3}=\dfrac{20}{3}\\\\x=0\to h(0)=\dfrac{1}{3}|0-6|+4=\dfrac{1}{3}|-6|+4=\dfrac{1}{3}(6)+4=2+4=6$

6 0

3 years ago

1) Fill in the Truth Table.

lakkis [162]

Answer:

kwjwkwkujdjdjwbw b2jwsidiwb

6 0

2 years ago

How to solve a mixed fraction?

Hunter-Best [27]

To solve a mixed fraction you will have to multiply the mixed number by the denominator then add the numerator for example:
7 3/4 you are gonna go 7x4=28 then you are gonna take that 28 and add 3 to it so it would be 31. I hope that helped some c:

5 0

3 years ago