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3 years ago

Create an equation with one solution of x = −4.

Mathematics

1 answer:

katen-ka-za [31]3 years ago

3 0

Answer:

What is missing is -5x - 20

Step-by-step explanation:

To find this, simply solve for the parenthesis as if it were a variable of its own. This will allow you to see exactly what is missing.

8x - (x - 20) = 2x - (_______)

8x - x + 20 = 2x - (_______)

7x + 20 = 2x - (_______)

5x + 20 = - (_______)

-5x - 20 = (_______)

So the answer for the parenthesis is -5x - 20

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3 years ago

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Let Y1 and Y2 be independent exponentially distributed random variables, each with mean 7. Find P(Y1 > Y2 | Y1 < 2Y2). (En

ArbitrLikvidat [17]

Y₁ and Y₂ are independent, so their joint density is

$f_{Y_1,Y_2}(y_1,y_2)=f_{Y_1}(y_1)f_{Y_2}(y_2)=\begin{cases}\frac1{49}e^{-\frac{y_1+y_2}7}&\text{for }y_1\ge0,y_2\ge0\\0&\text{otherwise}\end{cases}$

By definition of conditional probability,

P(Y₁ > Y₂ | Y₁ < 2 Y₂) = P((Y₁ > Y₂) and (Y₁ < 2 Y₂)) / P(Y₁ < 2 Y₂)

Use the joint density to compute the component probabilities:

• numerator:

$P((Y_1>Y_2)\text{ and }(Y_1$

$=\displaystyle\frac1{49}\int_0^\infty\int_{\frac{y_1}2}^{y_1}e^{-\frac{y_1+y_2}7}\,\mathrm dy_2\,\mathrm dy_1$

$=\displaystyle-\frac17\int_0^\infty\int_{-\frac{3y_1}{14}}^{-\frac{2y_1}7}e^u\,\mathrm du\,\mathrm dy_1$

$=\displaystyle-\frac17\int_0^\infty\left(e^{-\frac{2y_1}7} - e^{-\frac{3y_1}{14}}\right)\,\mathrm dy_1$

$=\displaystyle-\frac17\left(-\frac72e^{-\frac{2y_1}7} + \frac{14}3 e^{-\frac{3y_1}{14}}\right)\bigg|_0^\infty$

$=\displaystyle-\frac17\left(\frac72 - \frac{14}3\right)=\frac16$

• denominator:

$P(Y_1$

(I leave the details of the second integral to you)

Then you should end up with

P(Y₁ > Y₂ | Y₁ < 2 Y₂) = (1/6) / (2/3) = 1/4

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2 years ago

The probability of winning a lottery by selecting the correct six integers from the positive integers not exceeding 50 is _____

GREYUIT [131]

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8 0

2 years ago

PLEASE HELP ME If 0 < z ≤ 90 and sin(9z − 1) = cos(6z + 1), what is the value of z? z = 3 z = 4 z = 5 z = 6

Burka [1]

Answer:

z = 6

Step-by-step explanation:

We know that ...

sin(x) = cos(90 -x)

Substituting (9z-1) for x, this is ...

sin(9z -1) = cos(90 -(9z -1))

But we also are given ...

sin(9z -1) = cos(6z +1)

Equating the arguments of the cosine function, we have ...

90 -(9z -1) = 6z +1

90 = 15z . . . . . . . . . add (9z-1) to both sides

6 = z . . . . . . . . . . . . divide by 15

_____

Comment on the graph

The attached graph shows 5 solutions in the domain of interest. These come from the fact that the relation we used is actually ...

sin(x) = cos(90 +360k -x) . . . . . for any integer k

Then the above equation becomes ...

90 +360k = 15z

6 +24k = z . . . . . . . . . for any integer k

The sine and cosine functions also enjoy the relation ...

sin(x) = cos(x -90)

sin(9z -1) = cos(9z -1 -90) = cos(6z +1)

3z = 92 . . . . . equating arguments of cos( ) and adding 91-6z

z = 30 2/3

6 0

3 years ago