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3 years ago

What is the volume of a rectangular prism with dimensions 3ft 4ft and 12 ft?

Mathematics

2 answers:

ExtremeBDS [4]3 years ago

6 0

144

Explanation:3 times 4 is 12
12 times 12 is 144

andrew-mc [135]3 years ago

3 0

Answer:

144

Step-by-step explanation:

You might be interested in

Express the repeating decimal number as a quotient of two integers 2.63

arlik [135]

Answer: 2 63/100

(2 and 63 hundredths)

8 0

3 years ago

Read 2 more answers

Here are five number cards.

marta [7]

Possible ways to get >32:
19 , 14
14, 19
18,22
22, 19
13,22
22,13
14, 22
22,14

That is 8 ways
Total number of permutations of 2 cards from 5
= 5P2
= 5!/3!
= 20
So required probability
= 8/20
= 2/5.

6 0

2 years ago

A retired woman has $280,000 to invest. She has chosen one relatively safe investment fund that has an annual yield of 9% and an

ASHA 777 [7]

Answer:

9% fund: $ 210,000

13% fund: $70,000

Step-by-step explanation:

As she wants to have a $28,000 annual return for her $280,000 investment, she is expecting a return rate of 10%:

$r=\dfrac{R}{C}=\dfrac{28,000}{280,000}=0.10$

If we call x the proportion of the capital in the 9% fund, then (1-x) is the proportion of the capital in the 13% fund,and the return of the combination has to be the expected return of 10%:

$0.09x+0.13(1-x)=0.10\\\\0.09x+0.13-0.13x=0.10\\\\-0.04x=0.10-0.13=-0.03\\\\x=\dfrac{0.03}{0.04}=0.75$

Then, we know that 75% of the capital should be invested in the 9% fund and 25% in the 13% fund.

This correspond to a capital of:

9% fund: 0.75*$280,000 = $ 210,000

13% fund: 0.25*$280,000 = $70,000

6 0

4 years ago

There are 14 juniors and 16 seniors in a chess club. a) From the 30 members, how many ways are there to arrange 5 members of the

LiRa [457]

Answer:

No. of juniors = 14

No. of seniors = 16

Total students = 30

A) From the 30 members, how many ways are there to arrange 5 members of the club in a line?

Since we are asked about arrangement so we will use permutation

Formula : $^nP_r=\frac{n!}{(n-r)!}$

n = 30

r = 5

$^{30}P_5=\frac{30!}{(30-5)!}$

$^{30}P_5=17100720$

So, From the 30 members, there are 17100720 ways to arrange 5 members of the club in a line?

B) How many ways are there to arrange 5 members of the club in a line if there must be a senior at the beginning of the line and at the end of the line?

Out of 16 seniors 2 will be selected

So, 3 places are vacant

Remaining students = 30-2 = 28

So, out of 28 students 3 students will be selected

No. of ways = $^{16}P_2 \times ^{28}P_3$

No. of ways = $\frac{16!}{(16-2)!}\times\frac{28!}{(28-3)!}$

= $4717440$

There are 4717440 ways to arrange 5 members of the club in a line if there must be a senior at the beginning of the line and at the end of the line.

C)If the club sends 2 juniors and 2 seniors to the tournament, how many possible groupings are there?

Since we are not asked about arrangement so we will use combination

Out of 16 seniors 2 will be selected

Out of 14 juniors 2 will be selected

Formula : $^nC_r=\frac{n!}{r!(n-r)!}$