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2 years ago

Match the expressions with the property used to generate 5y + 2y + 6 + 2.

Mathematics

1 answer:

enyata [817]2 years ago

8 0

7y + 8 should be your answer.
Hope this helps!

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The middle-school camera club sold 240 tulip bulbs and 360 daffodil bulbs. Students divided the bulbs into 100 bags to sell at t

Travka [436]

240 bulbs + 360 bulbs = 600 bulbs
The 600 total bulbs would then be divided by 100 bags and you would get 6 bulbs per bag as an answer
Hope this works gohan! xD

3 0

2 years ago

Isabella can afford a $410-per-month car payment, and she's interested in either a sedan, which costs $21,600, or a station wago

mylen [45]

Answer:

isabella can afford neither the sedan nor the station wagon

Step-by-step explanation:

you read it and stop being a lazy

5 0

2 years ago

Read 2 more answers

The baseball team needs new equipment. Company A can provide 9 helmets, 6 bats, and 12 balls for $525. Company B can provide 10

Olin [163]

Answer:

Choice A.

Step-by-step explanation:

Let x = price of a helmet, y = price of a bat, z = price of a ball.

Company A can provide 9 helmets, 6 bats, and 12 balls for $525.

9x + 6y + 12z = 525

Company B can provide 10 helmets, 8 bats, and 10 balls for $600.

10x + 8y + 10z = 600

Company C can provide 8 helmets, 5 bats, and 15 balls for $500.

8x + 5y + 15z = 500

Answer: Choice A.

6 0

2 years ago

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Given the figure below is a special type of trapezoid and WX || YZ, which angle pairs can be proven supplementary by the given i

RSB [31]

Answer:

The answer:

<W and <Z

<X and <Y

<W and <X

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Probability extension: Jerry did only 5 problems of one assignment. What is the probability that the problems he did comprised t

zimovet [89]

Answer:

(b) 1/792

Step-by-step explanation:

The complete question is;

Counting: Grading One professor grades homework by randomly choosing 5 out of 12 homework problems to grade.

(a) How many different groups of 5 problems can be chosen from the 12

problems?

(b)Probability extension: Jerry did only 5 problems of one assignment. What is the probability that the problems he did comprised the group that was selected to be graded?

In (a)

Apply the formula

$\frac{n!}{(n-r)!(r!)}$

where n=12 and r=5

substitute values

$=\frac{12!}{(12-5)!(5!)} \\\\\\=\frac{12!}{(7!)(5!)} \\\\\\=\frac{12*11*10*9*8}{5*4*3*2*1} \\\\\\=\frac{95040}{120} \\\\\\=792$

In (b)

If Jerry did only 5 problems of one assignment then the probability will be

$\frac{5}{12} *\frac{4}{11} *\frac{3}{10} *\frac{2}{9} *\frac{1}{8} =\frac{1}{792}$

4 0

2 years ago