lol helpppppppppp ‍♀️

Answer:

(2, 5)

Step-by-step explanation:

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

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:

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:

22

Step-by-step explanation:

a+b^2

= -3+(5^2)

= -3+25

=22

Answer:

D

Step-by-step explanation:

The mean, median, and mode all are 3

Answer:

0.045 kg of cookies

Answer: 0.045

Step-by-step explanation:

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:

it's a bit blurry but if I aint wrong it's the 2nd one that say 2/5

Answer:

2.5 kilograms of chocolate

Step-by-step explanation:

Answer:

10×2+ 5÷2

=20+2.5

=22.5

Answer:

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

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:

Question 1: A

Question 2: D

Step-by-step explanation:

I hope it helps

Answer:

4,500

Step-by-step explanation:

just cause

Median of the given set of price is $4500

The median is the value separating the higher half from the lower half of a data sample, a population, or a probability distribution

Given set of data = $7000, $2000, $9000, $3000, $5000, $4000

Order the values from low to high

$2000, $3000, $4000, $5000, $7000, $9000

Find the middle term

Here number of terms = 6

It is even number  

Then find [tex]\frac{n}{2}[/tex] and [tex]\frac{n}{2}+1[/tex] terms

[tex]\frac{n}{2}=\frac{6}{2}=3[/tex]

[tex]\frac{n}{2}+1=3+1=4[/tex]

Median is the mean of the 3rd value and 4th value

Median =[tex]\frac{4000+5000}{2}=4500[/tex]

Hence, the median of the given set of price is $4500

Learn more about Median here

https://brainly.com/question/15923434

#SPJ2

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:

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:

G

Step-by-step explanation:

Answer:

its G

Step-by-step explanation:

Answer:

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

Answer:

4 x’s for 1/4

2 x’s for 2/4

1 x for 3/4

2 x’s for 1

Step-by-step explanation:

Answer:

Step-by-step explanation:

use your calculator

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:

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:

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:

(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:

B - obtuse

D - acute

Step-by-step explanation:

Answer:

I think angle B is an obtuse angle and D is an acute angle. I'm not sure.

Hope this helps :)

By the polynomial remainder theorem, the remainder is

f (2) = 2⁵ + 5×2³ + 4 = 32 + 40 + 4 = 76

(The theorem says a polynomial p(x) has remainder p(c) upon dividing it by x - c.)

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]