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ycow [4]
10 months ago
11

Y=−2x+4y, equals, minus, 2, x, plus, 4

Mathematics
1 answer:
vlabodo [156]10 months ago
3 0

The missing value of the solution to the equation y = −2x + 4, is: 3.

<h3>What is the Solution to an Equation?</h3>

The solution to a given equation, is the x and y values of an ordered pair that would make the equation true, if we substitute their values into the equation.

Given the equation, y = -2x + 4, and we have (x, -2) as the solution where the value of x is the missing value in the solution, to find the value of x, substitute y = -2 into the equation and solve for x:

-2 = -2x + 4

Subtract 4 from both sides

-2 - 4 = -2x + 4 - 4 (subtraction property of equality)

-6 = -2x

Divide both sides by -2

-6/-2 = -2x/-2

3 = x

x = 3

Therefore, the missing value is: 3.

Learn more about the solution to an equation on:

brainly.com/question/25678139

#SPJ1

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Let Y1 and Y2 have the joint probability density function given by:
Ann [662]

Answer:

a) k=6

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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

b)

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
Find the values of X and Y that makes these triangles congruent by the HL theorem
muminat

Answer:

C. x = 3, y = 2

Step-by-step explanation:

If both triangles are congruent by the HL Theorem, then their hypotenuse and a corresponding leg would be equal to each other.

Thus:

x + 3 = 3y (eqn. 1) => equal hypotenuse

Also,

x = y + 1 (eqn. 2) => equal legs

✔️Substitute x = y + 1 into eqn. 1 to find y.

x + 3 = 3y (eqn. 1)

(y + 1) + 3 = 3y

y + 1 + 3 = 3y

y + 4 = 3y

y + 4 - y = 3y - y

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Divide both sides by 2

4/2 = 2y/2

2 = y

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✔️ Substitute y = 2 into eqn. 2 to find x.

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8 0
3 years ago
Does 2 and 12 have a least common multiple of 24
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Let \mu be the population mean .

We are given that the mean potassium content of a popular sports drink is listed as 140 mg in a 32-oz bottle.

i.e. Null hypothesis :H_0:\mu=140

Alternative hypothesis for two tail hypothesis has sign (≠).

i.e. Alternative hypothesis : H_1 : \mu\neq140

∴ The hypotheses for a two-tailed test of the claimed potassium content:

H0: μ = 140 mg vs. H1: μ ≠ 140 mg

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