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2 years ago

Someone plz help me :(

Mathematics

1 answer:

butalik [34]2 years ago

6 0

You have it correct the answer is 20

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The value of a car purchased for $20,000 decreases at a rate of 12% per year. What will be the value of the car after 3 years?

DENIUS [597]

Answer:

The answer is A

Step-by-step explanation:

$20,000 • .12= $2,400

$2,400 • 3= $7,200

$20,000 - $7,200= $12,800

$12,800

4 0

3 years ago

when measuring time 8 concerts are approximately equal to 6 symphonies how many concerts are equal to 28 concerts how many conce

tatiyna

Lets answer this:-

1st lets divide to see how many times 28 goes in 8. And then we will multiply....

28 ÷ 8 = 3.5

Now lets multiply to see how many symphonies are equal to 25 concerts. Lets do it:-

3.5 × 6 = 21

So, 21 symphonies are equal to 28 concerts.

Lets divide to see how many concerts are equal to 42 symphonies and the multiply to get our final answer.

42 ÷ 6 = 7

7 × 8 = 56.

So, 56 concerts are equal to 42 symphonies.

Hope I helped ya!!

3 0

3 years ago

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Which two abiotic factors (also known as climate) determine the types of plants and animals that will be found in each terrestri

mafiozo [28]

The answer is A. Temperature and Precipitation since these are the most important factors that, for example, determine if the animal lives in a tropical or dry region, live closer to the poles, etc.

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3 years ago

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1)How many ways can the letters in the word BOOKKEEPER be arranged? (Must show set-up )

sergeinik [125]

Question 1

There are 5 letters (B, O, K, E, R) and there is a total of 10 letters to make up the word.
There are $\frac{10!}{(10-6)!6!}$ ways of arranging the letters, which equal to 210 ways

Question 2

There are seven swimmers in total.
There are $\frac{7!}{(7-1)!1!}$ ways of choosing the first winner, which is 7 ways
There are $\frac{6!}{(6-1)!1!}$ ways of choosing the second winner, which is 6 ways
There are $\frac{5!}{(5-1)!1!}$ ways of choosing the third winner, which is 5 ways
There are 7×6×5=210 ways of choosing first, second, and third winner

Question 3

The probability of eating an orange and a red candy is $\frac{15}{31}$ × $\frac{9}{30}$ , which equals to $\frac{9}{62}$

The probability of eating two green candies is $\frac{7}{31}$ × $\frac{6}{30}$ which equals to $\frac{7}{155}$

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3 years ago

Use the line of best fit to determine how many popsicles would be sold when the temperature is 100°.

SpyIntel [72]

Answer:

Step-by-step explanation:

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3 years ago

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