functions A and B are shown in the equation and graph. select the true statement. ​

Answer:

(-19/2, -5)

Step-by-step explanation:

Given the two equivalent equations

4x – 9y = 7 .... 1

–4x + 6y = 8 ....2

Adding both equations

-9y + 6y = 7+8

-3y = 15

y = -15/3

y = -5

Substitute y = -5 into 1

From 1; 4x - 9y = 7

4x - 9(-5) = 7

4x + 45 = 7

4x = 7 - 45

4x = -38

x = -38/4

x = -19/2

Hence the solution is (-19/2, -5)

Answer:

(2, 5)

Step-by-step explanation:

(x,y) -> )- y,x)

Answer:

I dont even know lol

Step-by-step explanation:

Answer:

Like when its in a file? They just like to take your points

Step-by-step explanation:

Answer:

15 C

Step-by-step explanation:

C = 5/9( F - 32)

Let F = 59

C = 5/9( 59 -32)

Parentheses first

   = 5/9( 27)

   = 15

Answer:

18 x 6 =108. she spent 108 minutes reading.

Step-by-step explanation:

Answer:

Question 1: A

Question 2: D

Step-by-step explanation:

I hope it helps

Answer:

Step-by-step explanation:

use your calculator

Answer:

D

Step-by-step explanation:

The mean, median, and mode all are 3

Answer:

10×2+ 5÷2

=20+2.5

=22.5

First, write the expression. Then solve.

9 x 10^-9

9 x 0.0000000010

0.000000009

---

sorry if i'm wrong, I haven't done this in a while

Answer:

219 percent

Step-by-step explanation:

97 goes into 213 two times which is a 200 percent increase which leaves 18 books. 18.5 percent of 97 is about 18 so rounded up it is a 219 percent increase

Answer:

0.045 kg of cookies

Answer: 0.045

Step-by-step explanation:

Answer:

(a) AH < HC is No

(b) AH < AC is Yes

(c) △AHC ≅ △AHB is Yes

Step-by-step explanation:

Given

See attachment for triangle

Solving (a): AH < HC

Line AH divides the triangle into two equal right-angled triangles which are: ABH and ACH (both right-angled at H).

To get the lengths of AH and HC, we need to first determine the measure of angles HAC and ACH. The largest of those angles will determine the longest of AH and HC. Since the measure of the angles are unknown, then we can not say for sure that AH < HC because the possible relationship between both lines are: AH < HC, AH = HC and AH > HC

Hence: AH < HC is No

Solving (b): AH < AC

Length AC represents the hypotenuse of triangle ACH, hence it is the longest length of ACH.

This means that:

AH < AC is Yes

Solving (c): △AHC ≅ △AHB

This has been addresed in (a);

Hence:

△AHC ≅ △AHB is Yes

Answer:

The mean absolute deviation is 2.25. Hope this helps

Step-by-step explanation:

2+4+6+9 divided by 4=5.25

5.25-2=3.25

5.25-4=1.25

6-5.25=0.75

9-5.25=3.75

3.25+1.25+0.75+3.75=9 divided by 4=2.25

Answer:

22

Step-by-step explanation:

a+b^2

= -3+(5^2)

= -3+25

=22

Answer:

m=8

Step-by-step explanation:

Divide each term in

7

m

=

56

7

m

=

56

by

7

7

.

7

m

7

=

56

7

7

m

7

=

56

7

Cancel the common factor of

7

7

.

Tap for more steps...

m

=

56

7

m

=

56

7

Divide

56

56

by

7

7

.

m

=

8

m

=

8

Step-by-step explanation:

7m=56

you have to have m by itself so to do that divide 7 on both sides.

7m/7=56/7

the 7 gets cancels on the left side so now you have

m=56/7

you divide

you get m=8

Answer:

0.7061 = 70.61% probability she will have her first crash within the first 30 races she runs this season

Step-by-step explanation:

For each race, there are only two possible outcomes. Either the person has a crash, or the person does not. The probability of having a crash during a race is independent of whether there was a crash in any other race. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

A certain performer has an independent .04 probability of a crash in each race.

This means that [tex]p = 0.04[/tex]

a) What is the probability she will have her first crash within the first 30 races she runs this season

This is:

[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]

When [tex]n = 30[/tex]

We have that:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{30,0}.(0.04)^{0}.(0.96)^{30} = 0.2939[/tex]

[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.2939 = 0.7061[/tex]

0.7061 = 70.61% probability she will have her first crash within the first 30 races she runs this season

Answer:

B and D

Step-by-step explanation:

This is because for B you do 8x9x1 which is basically 8x9 which is 72. Then, for D you do 12x3x2. Do 3x2 first and you get 6. Then, do 6x12 which is also 72. So the correct answers that will give you the volume of 72 units cube is B and D. Hope this helps!

Answer:

Feet and yards; the 1 whole strip could represent 1 yd. and the [tex]\frac{1}{3}[/tex] strip could represent feet.

Answer:

y=x+31.25

Because x stand for cauclator so 31.25 per x so it would end up being x+31.25

Step-by-step explanation:

Without replacement is very important

so u r already finished? don't post it next time

Answer:

[tex](a)\ g(x) = \frac{2}{3}(x+1)[/tex]

[tex](b)\ h(y) = \frac{1}{3}[1 + 4y][/tex]

[tex](c)[/tex] [tex]P(x>0.5) =\frac{5}{12}[/tex]

Step-by-step explanation:

Given

[tex]f(x,y) = \left \{ {{\frac{2}{3}(x+2y)\ \ 0\le x \le 1,\ 0\le y\le 1} \right.[/tex]

Solving (a): The marginal density of X

This is calculated as:

[tex]g(x) = \int\limits^{\infty}_{-\infty} {f(x,y)} \, dy[/tex]

[tex]g(x) = \int\limits^{1}_{0} {\frac{2}{3}(x + 2y)} \, dy[/tex]

[tex]g(x) = \frac{2}{3}\int\limits^{1}_{0} {(x + 2y)} \, dy[/tex]

Integrate

[tex]g(x) = \frac{2}{3}(xy+y^2)|\limits^{1}_{0}[/tex]

Substitute 1 and 0 for y

[tex]g(x) = \frac{2}{3}[(x*1+1^2) - (x*0 + 0^2)}[/tex]

[tex]g(x) = \frac{2}{3}[(x+1)}[/tex]

Solving (b): The marginal density of Y

This is calculated as:

[tex]h(y) = \int\limits^{\infty}_{-\infty} {f(x,y)} \, dx[/tex]

[tex]h(y) = \int\limits^{1}_{0} {\frac{2}{3}(x + 2y)} \, dx[/tex]

[tex]h(y) = \frac{2}{3}\int\limits^{1}_{0} {(x + 2y)} \, dx[/tex]

Integrate

[tex]h(y) = \frac{2}{3}(\frac{x^2}{2} + 2xy)|\limits^{1}_{0}[/tex]

Substitute 1 and 0 for x

[tex]h(y) = \frac{2}{3}[(\frac{1^2}{2} + 2y*1) - (\frac{0^2}{2} + 2y*0) ][/tex]

[tex]h(y) = \frac{2}{3}[(\frac{1}{2} + 2y)][/tex]

[tex]h(y) = \frac{1}{3}[1 + 4y][/tex]

Solving (c): The probability that the drive-through facility is busy less than one-half of the time.

This is represented as:

[tex]P(x>0.5)[/tex]

The solution is as follows:

[tex]P(x>0.5) = P(0\le x\le 0.5,0\le y\le 1)[/tex]

Represent as an integral

[tex]P(x>0.5) =\int\limits^1_0 \int\limits^{0.5}_0 {\frac{2}{3}(x + 2y)} \, dx dy[/tex]

[tex]P(x>0.5) =\frac{2}{3}\int\limits^1_0 \int\limits^{0.5}_0 {(x + 2y)} \, dx dy[/tex]

Integrate w.r.t. x

[tex]P(x>0.5) =\frac{2}{3}\int\limits^1_0 (\frac{x^2}{2} + 2xy) |^{0.5}_0\, dy[/tex]

[tex]P(x>0.5) =\frac{2}{3}\int\limits^1_0 [(\frac{0.5^2}{2} + 2*0.5y) -(\frac{0^2}{2} + 2*0y)], dy[/tex]

[tex]P(x>0.5) =\frac{2}{3}\int\limits^1_0 (0.125 + y), dy[/tex]

[tex]P(x>0.5) =\frac{2}{3}(0.125y + \frac{y^2}{2})|^{1}_{0}[/tex]

[tex]P(x>0.5) =\frac{2}{3}[(0.125*1 + \frac{1^2}{2}) - (0.125*0 + \frac{0^2}{2})][/tex]

[tex]P(x>0.5) =\frac{2}{3}[(0.125 + \frac{1}{2})][/tex]

[tex]P(x>0.5) =\frac{2}{3}[(0.125 + 0.5][/tex]

[tex]P(x>0.5) =\frac{2}{3} * 0.625[/tex]

[tex]P(x>0.5) =\frac{2 * 0.625}{3}[/tex]

[tex]P(x>0.5) =\frac{1.25}{3}[/tex]

Express as a fraction, properly

[tex]P(x>0.5) =\frac{1.25*4}{3*4}[/tex]

[tex]P(x>0.5) =\frac{5}{12}[/tex]

Answer:

Step-by-step explanation:

# 12

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

12 is a median of a data

Answer:

D

Step-by-step explanation:

Answer:D

Step-by-step explanation:

Answer:

G

Step-by-step explanation:

Answer:

its G

Step-by-step explanation: