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4 years ago

Select the answer choice that is equivalent to the expression given belon.

Mathematics

1 answer:

uranmaximum [27]4 years ago

6 0

Answer:

$23x - 10$

Step-by-step explanation:

To the find the equivalent of $9(2x - 3) + 5x + 17$ , evaluate the expression. Start by opening the bracket.

$9*2x - 9*3 + 5x + 17$

$18x - 27 + 5x + 17$

Pair like terms

$18x + 5x - 27 + 17$

$23x - 10$

The equivalent of $9(2x - 3) + 5x + 17$ is $23x - 10$

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Kelly drove north for 9 miles and then east for 12 miles at an average rate of 42 miles per hour to arrive at the town of Prime.

mr_godi [17]

Answer:

Step-by-step explanation:

This is one of the more interesting motion problems I've seen. I like it! If Kelly is driving north (straight up) for 9 miles, then turns east (right) and drives for 12 miles, what we have there are 2 sides of a right triangle. The hypotenuse is created by Brenda's trip, which originated from the same starting point as Kelly and went straight to the destination, no turns. We need the distance formula to solve this problem, so that means we need to find the distance that Brenda drove. Using Pythagorean's Theorem:

$9^2+12^2=c^2$ and

$81+144=c^2$ and

$225=c^2$ so

c = 15.

Brenda drove 15 miles. Now we can fill in a table with the info:

d = r x t

Kelly 12+9 42 t

Brenda 15 45 t

Because they both left at the same time, t represents that same time, whatever that time is. That's our unknown.

If d = rt, then for Kelly:

21 = 42t

For Brenda

15 = 45t

Solve Kelly's equation for t to get

t = 1/2 hr or 30 minutes

Solve Brenda's equations for t to get

t = 1/3 hr or 20 minutes

That means that Brenda arrived at the destination 10 minutes sooner than Kelly.

4 0

4 years ago

This makes no sense, math, i need help because this is BS and i don’t get it. there must be something i am missing.

nevsk [136]

Answer:

You are correct! But note the syntax:

the computer wants you to type in the box

2/3 x - 2

because "y =" is already written!

Step-by-step explanation:

4 0

3 years ago

Which could not be the number of tennis balls coach Kunal has ?

Nikitich [7]

2 one in his pocket and the one he's playing with.

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3 years ago

What is the meaning of {x} + {y}

Lelechka [254]

Are you talking about modulus functions?
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4 0

3 years ago

Read 2 more answers

Let Y1 and Y2 denote the proportion of time during which employees I and II actually performed their assigned tasks during a wor

Lemur [1.5K]

Answer:

Step-by-step explanation:

From the information given:

The joint density of $y_1$ and $y_2$ is given by:

$f_{(y_1,y_2)} \left \{ {{y_1+y_2, \ \ 0\ \le \ y_1 \ \le 1 , \ \ 0 \ \ \le y_2 \ \ \le 1} \atop {0, \ \ \ elsewhere \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \right.$

a)To find the marginal density of $y_1$ .

$f_{y_1} (y_1) = \int \limits ^{\infty}_{-\infty} f_{y_1,y_2} (y_1 >y_2) \ dy_2$

$=\int \limits ^{1}_{0}(y_1+y_2)\ dy_2$

$=\int \limits ^{1}_{0} \ \ y_1dy_2+ \int \limits ^{1}_{0} \ y_2 dy_2$

$= y_1 \ \int \limits ^{1}_{0} dy_2+ \int \limits ^{1}_{0} \ y_2 dy_2$

$= y_1[y_2]^1_0 + \bigg [ \dfrac{y_2^2}{2}\bigg]^1_0$

$= y_1 [1] + [\dfrac{1}{2}]$

$= y_1 + \dfrac{1}{2}$

i.e.

$f_{(y_1}(y_1)}= \left \{ {{y_1+\dfrac{1}{2}, \ \ 0\ \ \le \ y_1 \ \le , \ 1} \atop {0, \ \ \ elsewhere \ \\ \ \ \ \ \ \ \ \ } \right.$

The marginal density of $y_2$ is:

$f_{y_1} (y_2) = \int \limits ^{\infty}_{-\infty} fy_1y_1(y_1-y_2) dy_1$

$= \int \limits ^1_0 \ y_1 dy_1 + y_2 \int \limits ^1_0 dy_1$

$=\bigg[ \dfrac{y_1^2}{2} \bigg]^1_0 + y_2 [y_1]^1_0$

$= [ \dfrac{1}{2}] + y_2 [1]$

$= y_2 + \dfrac{1}{2}$

i.e.

$f_{(y_1}(y_2)}= \left \{ {{y_2+\dfrac{1}{2}, \ \ 0\ \ \le \ y_1 \ \le , \ 1} \atop {0, \ \ \ elsewhere \ \\ \ \ \ \ \ \ \ \ } \right.$

$P\bigg[y_1 \ge \dfrac{1}{2}\bigg |y_2 \ge \dfrac{1}{2} \bigg] = \dfrac{P\bigg [y_1 \ge \dfrac{1}{2} . y_2 \ge\dfrac{1}{2} \bigg]}{P\bigg[ y_2 \ge \dfrac{1}{2}\bigg]}$

$= \dfrac{\int \limits ^1_{\frac{1}{2}} \int \limits ^1_{\frac{1}{2}} f_{y_1,y_1(y_1-y_2) dy_1dy_2}}{\int \limits ^1_{\frac{1}{2}} fy_1 (y_2) \ dy_2}$

$= \dfrac{\int \limits ^1_{\frac{1}{2}} \int \limits ^1_{\frac{1}{2}} (y_1+y_2) \ dy_1 dy_2}{\int \limits ^1_{\frac{1}{2}} (y_2 + \dfrac{1}{2}) \ dy_2}$