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zmey [24]
3 years ago
15

A rectangular prism has a length of 14 cm, width of 2 cm, and height of 3 cm. What is the volume?

Mathematics
2 answers:
Cloud [144]3 years ago
4 0

Answer: 84 cubic cm

Step-by-step explanation: You just do 14 times 2 and the answer times 3 which is 84 cubic cm.

Ivenika [448]3 years ago
4 0

Answer: 84 cm³

Explanation: To find the volume of a rectangular prism or a prism or a prism whose base is a rectangle, we use the following formula.

Volume = length × width × height

Since the rectangular prism has a length of 14 cm, a width of 2 cm, and a height of 3cm, we cna plug this information into the formula to get

(14 cm)(2 cm)(3 cm).

(14)(2) is 28 and (28)(3) is 84 and notice that we have

centimeters × centimeters × centimeters or cm³.

So the volume of the given rectangular prism is 84 cm³.

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Can u guys PLEASE answer this question ASAP.
Andrej [43]

Answer:

  • 1. Interest: $166.67

           Principal: $6,166.67

  • 2. Interest: $200.00

            Principal: $5,200.00

Explanation:

<u><em>1. $6000 for 50 days at 20% p.a</em></u>

<u><em></em></u>

In 20% pa, pa means "per annum", i.e. "per year".

Assume simple interest:

  • Principal: $6,000
  • interest: 20% pa
  • time = 50 days

Interest:

  • Interest = Principal × number of days × annual rate / 360

  • Interest = $6,000 × 50 × 20% / 360 = $166.67

Principal = principal + interest = $6,000 + $166.67 = $6,166.67

<u><em></em></u>

<u><em>2. $5000 for 5 months at 0.8% per month</em></u>

Assume, again, simple interest.

  • Principal: $5,000
  • interest: 0.80% per month
  • time = 5 months

Interest:

  • Interest = Principal × number of months × montly rate

  • Interest = $5,000 × 5 × 0.80%  = $200.00

Principal = principal + interest = $5,000 + $200.00 = $5,200

You can see that the accrued interests depend on the principal, the interest rate, and the time.

8 0
3 years ago
It took you 120 minutes to work 15 math problems. How long did you spend on<br> each problem
Karolina [17]

Answer:

8 minutes

Step-by-step explanation:

120÷ 15= 8

yup thats pretty much it

5 0
2 years ago
Read 2 more answers
Wayne is using a waterproof covering to cover a storage locker that measures 4 ft wide, 10 ft tall, and 6 ft long. Which is the
Margaret [11]

Given the dimensions of locker box are length = 6 feet, width = 4 feet, and height = 10 feet.

Wayne wants to cover the box, so we need to find its total surface area. The box is in the shape of rectangular prism.

We know the formula for total surface area of rectangular prism is given as follows :-

Total \;Surface \;Area = 2*(lw + wh + hl)

T.S.A. = 2·(6 x 4 + 4 x 10 + 10 x 6) = 2·(24 + 40 + 60) = 2·(124) = 248 feet²

Surface Area = 248 squared feet

To cover the locker with waterproof covering, we needed to find the surface area of the box. As we multiplied two dimensions at a time in the formula, so the units are "squared feet".

Hence, option A is correct i.e. squared feet or ft².

5 0
2 years ago
$<br> Slope:<br> y-intercept<br> Equation:
liraira [26]
Slope is 4/3, y intercept is 2, equation is y=4/3x +2
7 0
2 years ago
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A source of information randomly generates symbols from a four letter alphabet {w, x, y, z }. The probability of each symbol is
koban [17]

The expected length of code for one encoded symbol is

\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha\ell_\alpha

where p_\alpha is the probability of picking the letter \alpha, and \ell_\alpha is the length of code needed to encode \alpha. p_\alpha is given to us, and we have

\begin{cases}\ell_w=1\\\ell_x=2\\\ell_y=\ell_z=3\end{cases}

so that we expect a contribution of

\dfrac12+\dfrac24+\dfrac{2\cdot3}8=\dfrac{11}8=1.375

bits to the code per encoded letter. For a string of length n, we would then expect E[L]=1.375n.

By definition of variance, we have

\mathrm{Var}[L]=E\left[(L-E[L])^2\right]=E[L^2]-E[L]^2

For a string consisting of one letter, we have

\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha{\ell_\alpha}^2=\dfrac12+\dfrac{2^2}4+\dfrac{2\cdot3^2}8=\dfrac{15}4

so that the variance for the length such a string is

\dfrac{15}4-\left(\dfrac{11}8\right)^2=\dfrac{119}{64}\approx1.859

"squared" bits per encoded letter. For a string of length n, we would get \mathrm{Var}[L]=1.859n.

5 0
2 years ago
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