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emmainna [20.7K]

2 years ago

14

a right rectangular prism has a length of 2 1/2 feet a width of 3 feet and a height of 1 1/2 ft unit cubes with side lengths of

1/2 foot are added to completely fill the prison with no space remaining what is the volume in cubic feet or the right rectangular prism

1 answer:

Annette [7]2 years ago

7 0

Given the original dimensions of:
2 1/2 ft, 3 ft and 1 1/2
After addition of 1/2 ft on both sides, the new dimensions will be:
3 1/2 ft, 4 ft, and 2 1/2 ft
thus the volume of the the prism will be:
Volume=length*width*height
V=3 1/2×4×2 1/2
V=35 ft³

Answer: 35 ft³

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A culture of bacteria has an initial population of 8300 bacteria and doubles every 10 hours. Using the formula Pt=P0*2^t/d , whe

tatiyna

Answer:

10,219 bacteria

Step-by-step explanation:

In this question, we are asked to calculate the population of bacteria in a culture given the formula for the population increase.

From the question, we know the formula to use is

Pt=Po*2^t/d

From the parameters in the question, we identify that Pt is ?, Po is 8300 bacteria , t is 3 hours and d is 10 hours which is the doubling time. Inserting these figures we have;

Pt = 8300 * 2^(3/10)

Pt = 8300 * 2^0.3

Pt = 10,218.5

Pt = 10,219 bacteria to the nearest whole number

8 0

2 years ago

Identify the perimeter and area of a square with diagonal length 24in. Give your answer in simplest radical form!!!

enyata [817]

Answer:

Perimeter: $48\sqrt{2} \: \mathrm{inches}$

Area: $288\mathrm \: \mathrm{square \: inches}$

Step-by-step explanation:

If the diagonal of the square is 24 inches, the side length of the square is $12\sqrt{2}$ inches ( $2x^2=24^2$ ). The perimeter of this square is then $4 \cdot 12\sqrt{2}=\fbox{$48\sqrt{2}\: \mathrm{in}$}$ and the area $(12\sqrt{2})^2=\fbox{$288\: \mathrm{in^2}$}$ .

5 0

1 year ago

PLEASE HELP! I WILL GIVE BRAINLYEST TO WHOEVER ANSWERS FIRST! PLEASE PLEASE BE QUICK! thank you! and please include explanation

katrin2010 [14]

Answer:

A=920 ft.

Step-by-step explanation:

Using the formulas

A=2AB+(a+b+c)h

AB=s(s﹣a)(s﹣b)(s﹣c)

s=a+b+c

2

Solving for A

A=ah+bh+ch+1

2﹣a4+2(ab)2+2(ac)2﹣b4+2(bc)2﹣c4=8·20+17·20+15·20+1

2·﹣84+2·(8·17)2+2·(8·15)2﹣174+2·(17·15)2﹣154=920

4 0

2 years ago

zheka24 [161]

The trick is to divide the numeral by the out of number and Multiply it by 100 ie 6/28 multiplied by 100 = 21,43%

3 0

2 years ago

Read 2 more answers

A 52-card deck is thoroughly shuffled and you are dealt a hand of 13 cards. (a) If you have at least one ace, what is the probab

jasenka [17]

Answer:

a) 0.371

b) 0.561

Step-by-step explanation:

We can answer both questions using conditional probability.

(a) We need to calculate the probability of obtaining two aces given that you obtained at least one. Let's call <em>A</em> the random variable that determines how many Aces you have. A is a discrete variable that can take any integer value from 0 to 4. We need to calculate

$P(A \geq 2 | A \geq 1) = P(A\geq 2 \cap A \geq 1) / P(A \geq 1)$

Since having 2 or more aces implies having at least one, the event $A \geq 2 \cap A \geq 1$ is equal to the event $A \geq 2$ . Therefore, we can rewrite the previous expression as follows

$P(A \geq 2) / P(A \geq 1)$

We can calculate each of the probabilities by substracting from one the probability of its complementary event, which are easier to compute

$P(A \geq 2) = 1 - P((A \geq 2)^c) = 1 - P((A = 0) \bigsqcup (A = 1)) = 1 - P(A = 0) - P (A = 1)$

$P (A \geq 1) = 1 - P ((A \geq 1)^c) = 1 - P(A = 0)$

We have now to calculate P(A = 0) and P(A = 1).

For the event A = 0, we have to pick 13 cards and obtain no ace at all. Since there are 4 aces on the deck, we need to pick 13 cards from a specific group of 48. The total of favourable cases is equivalent to the ammount of subsets of 13 elements of a set of 48, in other words it is $48 \choose 13$ . The total of cases is $52 \choose 13$ . We obtain

$P(A = 0) = {48 \choose 13}/{52 \choose 13} = \frac{48! * 39!}{52!*35!} \simeq 0.303$

For the event A = 1, we pick an Ace first, then we pick 12 cards that are no aces. Since we can pick from 4 aces, that would multiply the favourable cases by 4, so we conclude

$P(A=1) = 4*{48 \choose 12}/{52 \choose 13} = \frac{4*13*48! * 39!}{52!*36!} \simeq 0.438$

Hence,

$1 - P(A = 1)-P(A=0) /1-P(A=1) = 1 - 0.438 - 0.303/1-0.303 = 0.371$

We conclude that the probability of having two aces provided we have one is 0.371

b) For this problem, since we are guaranteed to obtain the ace of spades, we can concentrate on the other 12 cards instead. Those 12 cards have to contain at least one ace (other that the ace of spades).

We can interpret this problem as if we would have removed the ace of spades from the deck and we are dealt 12 cards instead of 13. We need at least one of the 3 remaining aces. We will use the random variable B defined by the amount of aces we have other that the ace of spades. We have to calculate the probability of B being greater or equal than 1. In order to calculate that we can compute the probability of the <em>complementary set</em> and substract that number from 1.

$P(B \geq 1) = 1-P(B=0)$

In order to calculate P(B=0), we consider the number of favourable cases in which we dont have aces. That number is equal to the amount of subsets of 12 elements from a set with 48 (the deck without aces). Then, the amount of favourable cases is $48 \choose 12$ . Without the ace of spades, we have 51 cards on the deck, therefore

$P(B = 0) = {48 \choose 12} / {51 \choose 12} = \frac{48!*39!}{51!*36!} = 0.438$

We can conclude

$P(B \geq 1) = 1- 0.438 = 0.561$

The probability to obtain at least 2 aces if we have the ace of spades is 0.561

4 0

2 years ago