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3 years ago

Please Help me solve: 7/8 * 1/4 Please give your answer in simplest form, Thank You!

Mathematics

1 answer:

alukav5142 [94]3 years ago

7 0

Make them the same demonatator so 7/8 times 2/8 which equals 14 over 8 for a mixed number of 1 6/8

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Which of these is the longest 1 foot 1 yard 3 feet 39 inches 2.5 feet

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39 inches is the longest bc a foot is 12 inches and the longest feet the is 3 feet or a yard. So 12 x 3 = 36 and that is 3 inches short than 39 so 39 is the longest.

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3 years ago

The goal of a benefit concert is to collect 50,000 cans of food. Food is collected at each entrance to the concert.

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Answer: They collected 13,204 cans more than needed.

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2 years ago

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What is 9 times -11 + 17 times 11 in factored form

Rama09 [41]

Answer:

Step-by-step explanation:

(17 * 11 = 187) + (9 * -11 = -99)

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3 years ago

I wish to have $10,000 at the end of 8 yers in a bank offering a simple interest rate of 7.5% per year. How much should I deposi

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Answer:

You have to invest $6,250

Step-by-step explanation:

very simple applying the simple interest formula which is

$A = P (1 + rt)$

Given data

A, final amount = $10,000

P, initial principal balance= ?

r, annual interest rate = 7.5%

t, time (in years)= 8 years

we can substitute our given data to find the principal needed.

$10000= P(1+0.075*8)\\\10000= P(1.6)\\\$

Divide both sides by 1.6 we have

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3 years ago

a. Among a shipment of 5,000 tires, 1,000 are slightly blemished. If one purchases 10 of these tires, what is the probability th

hodyreva [135]

Answer:

<h2>See the explanation.</h2>

Step-by-step explanation:

3 or less than 3 tires among the 10 tires should be blemished.

There are some possibilities.

1000 tires are blemished, (5000 - 1000) = 4000 tires are not blemished.

From the 5000 tires, 10 tires can be chosen in $^{5000}C_{10}$ ways.

Possibility 1: 3 tires are blemished.

Total 10 tires can be chosen with 3 blemished tires in $^{1000}C_3 \times {4000}C_7$ ways.

Possibility 2: 2 tires are blemished.

Total 10 tires can be chosen with 2 blemished tires in $^{1000}C_2 \times ^{4000}C_8$ ways.

Possibility 3: 1 tires are blemished.

Total 10 tires can be chosen with 1 blemished tires in $^{1000}C_1 \times^{4000}C_9$ ways.

Possibility 4: No tires are blemished.

Total 10 tires can be chosen with no blemished tires in $^{1000}C_0 \times^{4000}C_{10}$ ways.

Hence, the required probability is $\frac{^{1000}C_1 \times^{4000}C_9 + ^{1000}C_0 \times^{4000}C_{10} + ^{1000}C_2 \times^{4000}C_8 + ^{1000}C_3 \times^{4000}C_7}{^{5000}C_{10} }$ .

6 0

2 years ago