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leonid [27]
3 years ago
11

What is the volume enclosed by the slanted prism in the diagram?

Mathematics
2 answers:
Gwar [14]3 years ago
8 0
The answer is 225 cm3.
 
lbvjy [14]3 years ago
3 0
The answer is letter c. The volume enclosed by the slanted prism is 300 cm3. 

The formula of volume for any prism is as follows:
area of base *  height = volume

area = 25 cm2
height = 12 cm

25 cm2 * 12 cm = volume
volume = 300 cm3

Thank you for posting your question. Please feel free to ask me more.
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Answer:

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Step-by-step explanation:

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5 0
3 years ago
ABCD is a parallelogram. If AB = 3x and CD = x+10, find AB
Readme [11.4K]

A parallelogram by definition is a four-sided figure with opposite sides that are parallel.

If that's the case, the length of opposite sides should be the same as well, meaning that:

AB=CD

then,

\begin{gathered} 3x=x+10 \\ \text{solve for x} \\ 3x-x=10 \\ 2x=10 \\ x=\frac{10}{2} \\ x=5 \end{gathered}

find AB

\begin{gathered} AB=3x \\ AB=3\cdot5 \\ AB=15 \end{gathered}

A

7 0
1 year ago
During the period of time that a local university takes phone-in registrations, calls come in
Zinaida [17]

Answer:

a) The expected number of calls in one hour is 30.

b) There is a 21.38% probability of three calls in five minutes.

c) There is an 8.2% probability of no calls in a five minute period.

Step-by-step explanation:

In problems that we only have the mean during a time period can be solved by the Poisson probability distribution.

Poisson probability distribution

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

a. What is the expected number of calls in one hour?

Calls come in at the rate of one each two minutes. There are 60 minutes in one hour. This means that the expected number of calls in one hour is 30.

b. What is the probability of three calls in five minutes?

Calls come in at the rate of one each two minutes. So in five minutes, 2.5 calls are expected, which means that \mu = 2.5. We want to find P(X = 3).

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 3) = \frac{e^{-2.5}*(2.5)^{3}}{(3)!} = 0.2138

There is a 21.38% probability of three calls in five minutes.

c. What is the probability of no calls in a five-minute period?

This is P(X = 0) with \mu = 2.5.

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-2.5}*(2.5)^{0}}{(0)!} = 0.0820

There is an 8.2% probability of no calls in a five minute period.

6 0
3 years ago
Write 3 + 2 x 1/10 +4 x 1/1000 in standard form
maksim [4K]
The answer is 209.708
7 0
3 years ago
In a certain sequence of all positive terms, {a1, a2, a3, …} each term equals the previous term times a constant factor. If (a1)
andriy [413]

Let k be the constant factor between terms, so that

a_2=ka_1

a_3=ka_2=k^2a_1

and so on, with

a_n=ka_{n-1}=\cdots=k^{n-1}a_1

(notice how the exponent and the subscript add to n)

Then

a_5=k^4a_1

so if

a_1a_5=900

then

k^4{a_1}^2=900\implies k^4=\dfrac{900}{{a_1}^2}=\left(\dfrac{30}{a_1}\right)^2\implies k^2=\dfrac{30}{a_1}

Now,

a_3=k^2a_1=\dfrac{30a_1}{a_1}=30

so the third term in the sequence is 30.

7 0
3 years ago
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