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alexira [117]
3 years ago
12

Give three real-world examples of rectangular prisms. Then give two real-world examples of triangular prisms. Explain how you kn

ow the object you chose represents a prism.​
Mathematics
2 answers:
Vikki [24]3 years ago
4 0

Answer:

Sample response: Boxes, ice cubes, and a brick are examples of rectangular prisms. Ramps and tents are examples of triangular prisms. A rectangular prism has six rectangular faces. A triangular prism has two triangular faces and three rectangular faces.

Step-by-step explanation:

i tried

denpristay [2]3 years ago
3 0

Three real world examples of rectangular prisms include juice boxes, cereal boxes, and even cargo containers. Two real world examples of triangular prisms include camping tents and triangular roofs. I chose these objects to represent triangular and rectangular prisms because triangular prisms have two triangular faces and three rectangular faces and rectangular prisms have six rectangular faces.

Sample Response: Boxes, ice cubes, and brick are examples of rectangular prisms. Ramps and tents are examples of triangular prisms. A rectangular prism has six rectangular faces. A triangular prism has two triangular faces and three rectangular faces.

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Find the quotient. 12a^3p^4 ÷ -2a^2p
777dan777 [17]

Answer:

-(6ap³)

Step-by-step explanation:

We have to find the quotients of

\frac{12a^{3}p^{4}}{-2a^{2}p}

We take the common 2a² p from the numerator

= \frac{(6ap^{3}(2a^{2}p))}{(-2a^{2}p)}

= -(6ap³)

Therefore, the quotient after division will be -(6ap³)

4 0
3 years ago
How to divide fractions with whole numbers?
ad-work [718]
<span>1.Multiply the bottom number of the fraction by the whole number.
<span>2.Simplify the fraction (if needed)</span></span>
7 0
3 years ago
Which of the following shows the division problem??
elena-14-01-66 [18.8K]

we are given

\frac{3x^2-4x+9}{x-2}

we can see that

numerator is

3x^2-4x+9

and denominator is

x-2

We always put coefficient of numerator inside box

we can see that coefficient of numerators are 3 , -4 and 9

and in the left side , we put value of x after putting denominator =0

x-2=0

x=2

so, left side will be 2

so, option-D..................Answer

7 0
3 years ago
Last week f40 fifth graders went on a field trip to the science museum
Xelga [282]
8 out of 40 fifth graders went to see the lab animal presentation. This can be written as this expression:

\frac{8}{40}

Find the greatest common factor of 8 and 40:

Factors of 8: 1, 2, 4, 8
Factors of 40: 1, 2, 4, 5, 8, 10, 20, 40

Therefore the greatest common factor is 8. Divide both the numerator and the denominator by 8.

\frac{8}{40} = \frac{8/8}{40/8} = \frac{1}{5}

Therefore the fractional part is \frac{1}{5}.

To get the percentage, you need to get the denominator to a value of 100.

To do so, you need to find the value to multiply both the numerator and denominator by to achieve this.

To get this value, you can divide 100 by the denominator of the fractional value.

100 / 5 = 20

Then you can multiply both the numerator and denominator by 20:

\frac{1}{5} = \frac{1 * 20}{5 * 20} = \frac{20}{100}

From here, the numerator will state the percentage:

20\%

Hope this helps :)


8 0
3 years ago
Read 2 more answers
A shop sells a particular of video recorder. Assuming that the weekly demand for the video recorder is a Poisson variable with t
julia-pushkina [17]

Answer:

a) 0.5768 = 57.68% probability that the shop sells at least 3 in a week.

b) 0.988 = 98.8% probability that the shop sells at most 7 in a week.

c) 0.0104 = 1.04% probability that the shop sells more than 20 in a month.

Step-by-step explanation:

For questions a and b, the Poisson distribution is used, while for question c, the normal approximation is used.

Poisson 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^{-\lambda}*\lambda^{x}}{(x)!}

In which

x is the number of successes

e = 2.71828 is the Euler number

\lambda is the mean in the given interval.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The Poisson distribution can be approximated to the normal with \mu = \lambda, \sigma = \sqrt{\lambda}, if \lambda>10.

Poisson variable with the mean 3

This means that \lambda= 3.

(a) At least 3 in a week.

This is P(X \geq 3). So

P(X \geq 3) = 1 - P(X < 3)

In which:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

Then

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

P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498

P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494

P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240

So

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0498 + 0.1494 + 0.2240 = 0.4232

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 1 - 0.4232 = 0.5768

0.5768 = 57.68% probability that the shop sells at least 3 in a week.

(b) At most 7 in a week.

This is:

P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)

In which

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

P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498

P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494

P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240

P(X = 3) = \frac{e^{-3}*3^{3}}{(3)!} = 0.2240

P(X = 4) = \frac{e^{-3}*3^{4}}{(4)!} = 0.1680

P(X = 5) = \frac{e^{-3}*3^{5}}{(5)!} = 0.1008

P(X = 6) = \frac{e^{-3}*3^{6}}{(6)!} = 0.0504

P(X = 7) = \frac{e^{-3}*3^{7}}{(7)!} = 0.0216

Then

P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.0498 + 0.1494 + 0.2240 + 0.2240 + 0.1680 + 0.1008 + 0.0504 + 0.0216 = 0.988

0.988 = 98.8% probability that the shop sells at most 7 in a week.

(c) More than 20 in a month (4 weeks).

4 weeks, so:

\mu = \lambda = 4(3) = 12

\sigma = \sqrt{\lambda} = \sqrt{12}

The probability, using continuity correction, is P(X > 20 + 0.5) = P(X > 20.5), which is 1 subtracted by the p-value of Z when X = 20.5.

Z = \frac{X - \mu}{\sigma}

Z = \frac{20 - 12}{\sqrt{12}}

Z = 2.31

Z = 2.31 has a p-value of 0.9896.

1 - 0.9896 = 0.0104

0.0104 = 1.04% probability that the shop sells more than 20 in a month.

5 0
3 years ago
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