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3 years ago

Help please! will give brainliest

Mathematics

1 answer:

elena-14-01-66 [18.8K]3 years ago

7 0

Answer:

exact volume= 12320ft³

approximate volume= 12308.8ft³

Step-by-step explanation:

exact volume= 22/7×14² x 20

approximate volume = 3.14x14²x20

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Easy points

Finger [1]

Their brother will be 67 years old because when they were six their brother was 3. They were three years apart.

3 0

2 years ago

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Determine the present value P you must invest to have the future value A at simple intrest rate r after time t. A=$14,000, r=9.5

Ivan

Answer:

$10,144.93

Step-by-step explanation:

Amount formula; A = (Principal* rate* time) +Principal

A= (P*r*t) +P

A= 14,000

r= 9.5% OR 0.095 as a decimal

t= 4

Plug the numbers into the formula;

14,000 = (P* 0.095 *4 ) +P

14,000 = 0.38P +P

14,000 = 1.38P

Divide both sides by 1.38 to solve for P;

14,000/1.38 = P

P= 10,144.9275

Therefore, the present value (Principal) = $10,144.93

3 0

3 years ago

Simplify the expression. HELP PLEASE!!

ddd [48]

$\sqrt[2]{81} + \sqrt[3]{64} = 9 + 4 = 13

Answer : 13$

8 0

3 years ago

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If on a scale 75 feet is represented by 1 inch, then a scale of 1/5 inch represents how many feet?

lara [203]

Answer:

15 feet

Step-by-step explanation:

if 1 inch represents 75 and it 1/5 then you do 75 divided by 5 and it gives you 15 so it represents 15 feet

8 0

3 years ago

Five cards are drawn from a standard 52-card playing deck. A gambler has been dealt five cards—two aces, one king, one 3, and on

Nookie1986 [14]

Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

If he is given with two kings.
If he is given one king and one ace.

Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by = $\frac{^{3}C_2 }{^{47}C_2}$ {∵ one king is already there}

= $\frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}$ {∵ $^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{3}{1081}$ = 0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by = $\frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}$

= $\frac{3!}{1! \times 2!}\times \frac{2!}{1! \times 1!} \times \frac{2! \times 45!}{47!}$

= $\frac{6}{1081}$ = 0.0055

Now, probability that he ends up with a full house = $\frac{3}{1081} + \frac{6}{1081}$

= $\frac{9}{1081}$ = <u>0.0083</u>.

3 0

3 years ago

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