Help pleeeeeeeeeease ?

Answers

Answer 1

Answer: The answer is choice D.
Cause it can't be calculate.

Answer 2

Answer: D is irrational because the numbers keep on going on and on and on. I hope this helps!

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Solve. Q - 7 5/6 = 6 1/2

Answers

$Q - 7 (5)/(6) = 6 (1)/(2) \n \n Q - 7 (5)/(6) + 7 (5)/(6) = 6 (1)/(2) + 7 (5)/(6) \n \n Q = 6 (1*3)/(2*3) + 7 (5)/(6) \n \n Q = 6 + 7 + (3)/(6) + (5)/(6) \n \n Q = 13 + (8)/(6) = 13 + (6)/(6 ) + (2)/(6) = 14 (2)/(6 ) = 14 (1)/(3)$

At a high school, students can choose between three art electives, four history electives, and five computer electives. Each student can choose two electives. Which expression represents the probability that a student chooses an art elective and a history elective?

Answers

Answer:

$(^3C_1* ^4C_1)/(^(12)C_2)$

Step-by-step explanation:

Given,

Art electives = 3,

History electives = 4,

Computer electives = 5,

Total number of electives = 3 + 4 + 5 = 12,

Since, if a student chooses an art elective and a history elective,

So, the total combination of choosing an art elective and a history elective = $^3C_1* ^4C_1$

Also, the total combination of choosing any 2 subjects out of 12 subjects = $^(12)C_2$

Hence, the probability that a student chooses an art elective and a history elective = $\frac{\text{Total combination of choosing an art elective and a history}}{\text{ Total combination of choosing any 2 subjects}}$

$=(^3C_1* ^4C_1)/(^(12)C_2)$

Which is the required expression.

Answer: Hello!

we have:

3 art electives

4 history electives

5 computer electives

which adds to a total of 12.

If the selection is random, each elective has the same probability.

The probability of selecting an art electives is the quotient between the number of art electives and the total number of electives:

3/12

suppose that this event is true, now we need to see the probability of choosing also a history elective;

We do the same process as before, we have 4 history electives and, because we already selected 1 in the previous step, we have a total of 11 electives:

the probability now is 4/11.

Now we want to calculate the joint probability of bot events is equal to the product of their probabilities; this is:

p= (3/12)*(4/12) = (4*3)/(11*12) = 12/(11*12) = 1/11

But there is also the case where the selection is in the other order (first history and second art) so the probability is equal to

2*1/11 = 2/11

Simplify the following expression
2x − 6y + 3x2 + 7y − 14x

Answers

2x - 6y + 3x² + 7y - 14y = 3x² + (2x - 14x) + (-6y + 7y) = 3x² - 12x + y

Sweatshirt = $32 t-shirts = $14 ,Which costs more, 14 sweatshirts or 32 t-shirts?What property can be used to find the answer.

Answers

Cost of a sweatshirt = $32
Then
Cost of 14 sweatshirts = (32 * 14) dollars
= 448 dollars
Cost of a t-shirt = $14
Cost of 32 t-shirts = (32 * 14) dollars
= 448 dollars
So it can seen from the deduction that the cost of 14 sweatshirts and 32 t-shirts comes out to be same and equal to $448. I hope the answer and the procedure of doing this problem is clear to you. It is important to look at all the details given in the question and then start solving the problem.

14 sweatshirts would cost:
14 sweatshirts* ($32/ 1 sweatshirt)= $448.
(Note that the unit cancels out, so you get the answer)

32 T-shirts would cost:
32 T-shirts* ($14/ 1 T-shirt)= $448.
(Note that the unit cancels out, so you get the answer)

The cost of 14 sweatshirts and 32 T-shirts is equal~

Can someone plz help me with the bottom part

Answers

9.

2x +3 + x +9 = 180 => 3x + 12 = 180 => 3x = 168 => x = 56 => m(<ABD) = 2*56 +3 = 115.

At a carnival, food tickets cost $2 each and ride tickets cost $3 each. A total of $1,240 was collected at the carnival. The number of food tickets sold was 10 less than twice the number of ride tickets sold.The system of equations represents x, the number of food tickets sold, and y, the number of ride tickets sold.

2x + 3y = 1240

x = 2y – 10

How many of each type of ticket were sold?

180 food tickets and 293 ride tickets
180 food tickets and 350 ride tickets
293 food tickets and 180 ride tickets
350 food tickets and 180 ride tickets

Answers

The number of food tickets sold is 180 and ride tickets is 350.

In order to determine how many of each type of ticket was sold, the two equations given have to be solved simultaneously using the substitution method.

2x + 3y = 1240 equation 1

x = 2y – 10 equation 2

Substitute for x in equation 1

2(2y -10) + 3y = 1240

4y - 20 + 3y = 1240

Combine similar terms

7y = 1240 + 20

7y = 1260

Divide both sides of the equation by 7

y = 1260 / 7

y = 180

Substitute for y in equation 1

2x + 3(180) = 1240

2x + 540 = 1240

2x = 1240 - 540

2x = 700

x = 700 / 2

x = 350

To learn more about simultaneous equations, please check: brainly.com/question/25875552

Answer:

180 food tickets and 350 ride tickets

Step-by-step explanation:

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