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tamaranim1 [39]
4 years ago
15

A rectangular prism has a length of 8 in., a width of 4 in., and a height of 2 1/4 in. The prism is filled with cubes that have

edge lengths of 1/4 in. How many cubes are needed to fill the rectangular prism? Enter your answer in the box. To fill the rectangular prism, cubes are needed.
Mathematics
1 answer:
wariber [46]4 years ago
7 0
The prism is (8 in)/(1/4 in/cube) = 32 cubes by (4 in)/(1/4 in/cube) = 16 cubes by (9/4 in)/(1/4 in/cube) = 9 cubes.

The number of cubes needed to fill the prism is
.. 32 * 16 * 9 = 4608
You might be interested in
What 3 numbers multiply to get 63?
masya89 [10]
There are many combinations if the numbers are not required to be integers.

If they are required to be integers, I'd suggest:

3*7*3 = 63
5 0
3 years ago
Read 2 more answers
Use a=6 and b=24 to work out the value of these expressions
Reika [66]
A) 30
B) 144
C) 4
D) 900
3 0
3 years ago
A large pool of adults earning their first driver’s license includes 50% low-risk drivers, 30% moderate-risk drivers, and 20% hi
Mamont248 [21]

Answer:

The probability that these four will contain at least two more high-risk drivers than low-risk drivers is 0.0488.

Step-by-step explanation:

Denote the different kinds of drivers as follows:

L = low-risk drivers

M = moderate-risk drivers

H = high-risk drivers

The information provided is:

P (L) = 0.50

P (M) = 0.30

P (H) = 0.20

Now, it given that the insurance company writes four new policies for adults earning their first driver’s license.

The combination of 4 new drivers that satisfy the condition that there are at least two more high-risk drivers than low-risk drivers is:

S = {HHHH, HHHL, HHHM, HHMM}

Compute the probability of the combination {HHHH} as follows:

P (HHHH) = [P (H)]⁴

                = [0.20]⁴

                = 0.0016

Compute the probability of the combination {HHHL} as follows:

P (HHHL) = {4\choose 1} × [P (H)]³ × P (L)

               = 4 × (0.20)³ × 0.50

               = 0.016

Compute the probability of the combination {HHHM} as follows:

P (HHHL) = {4\choose 1} × [P (H)]³ × P (M)

               = 4 × (0.20)³ × 0.30

               = 0.0096

Compute the probability of the combination {HHMM} as follows:

P (HHMM) = {4\choose 2} × [P (H)]² × [P (M)]²

                 = 6 × (0.20)² × (0.30)²

                 = 0.0216

Then the probability that these four will contain at least two more high-risk drivers than low-risk drivers is:

P (at least two more H than L) = P (HHHH) + P (HHHL) + P (HHHM)

                                                            + P (HHMM)

                                                  = 0.0016 + 0.016 + 0.0096 + 0.0216

                                                  = 0.0488

Thus, the probability that these four will contain at least two more high-risk drivers than low-risk drivers is 0.0488.

6 0
4 years ago
Find the inverse Laplace transform f(t) of the function F(s). Write uc for the Heaviside function that turns on at c, not uc(t).
zzz [600]

Answer:

F(t)=\frac{-1}{2}e^{7(t-7)}+\frac{1}{2}e^{-7(t-7)}

Step-by-step explanation:

We have given F(S)=\frac{7e^{-7s}}{s^2-49}

Now  F(S)=e^{-7s}G(s)

Here G(S)=\frac{7}{S^2-49}

Now first find the Laplace inverse of G(S)

Using partial fraction

\frac{7}{(s+7)(s-7)}=\frac{A}{(S+7)}+\frac{B}{S-7}

7=A(S-7)+B(S+7)

On comparing the coefficient

A=\frac{1}{2}  and B=\frac{-1}{2}  

On putting the value of A and B  

G(S)=\frac{-1}{2(S+7)}+\frac{1}{2(S+7)}

Taking inverse Laplace

G(t)=\frac{-1}{2}e^{7t}+\frac{1}{2}e^{-7t}

Now in G(s) there is onether term e^{-7s}

So F(t)=\frac{-1}{2}e^{7(t-7)}+\frac{1}{2}e^{-7(t-7)}

6 0
3 years ago
**MARKING BRAINLIEST PLEASE HELP**
Annette [7]

Answer:

6

1, 50.3

Step-by-step explanation:

the second part for question two could be anywhere from 50.1-50.4

3 0
3 years ago
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