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3 years ago

A rectangular prism has a length of 6cm width of 3cm and a height of 4 1/2 cm.

Mathematics

2 answers:

lapo4ka [179]3 years ago

6 0

I guess it's <br />40.5 hope soo

balu736 [363]3 years ago

6 0

Answer:

To fill the rectangular prism, 648 cubes are needed.

Step-by-step explanation:

It is given that a rectangular prism has a length of 6 cm width of 3 cm and a height of 4 1/2 cm.

The volume of a rectangular prism is

$V=length\times breadth \times height$

$V_1=6\times 3 \times 4\frac{1}{2}$

$V_1=18 \times \frac{9}{2}$

$V_1=81$

The volume of rectangular prism is 81 cm³.

The volume of a cube is

$V=a^3$

Where a is the side length.

The volume of a cube with edge 1/2.

$V_2=(\frac{1}{2})^3$

$V_2=0.125$

To fill the rectangular prism, the required number of cubes is

$n=\frac{V_1}{V_2}$

$n=\frac{81}{0.125}$

$n=648$

Therefore, to fill the rectangular prism, 648 cubes are needed.

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3 years ago

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3 years ago

Read 2 more answers

A bus travels on an east-west highway connecting two cities A and B that are 100 miles apart. There are 2 services stations alon

melamori03 [73]

Answer:

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Step-by-step explanation:

To begin with you have to understand what is the distribution of the random variable. If X represents the point where the bus breaks down. That is correct.

X~ Uniform(0,100)

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$f(x) = P(X=x) = 1/100 \,\,\,\, \text{if} \,\,\,\, 0 \leq x \leq 100\\f(x) = P(X=x) = 0 \,\,\,\, \text{otherwise}$

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$D(x) =\left\{ \begin{array}{ll} x & \mbox{if } 0\leq x \leq 20 \\ x-20 & \mbox{if } 20\leq x < 45\\ 70-x & \mbox{if } 45 \leq x \leq 70\\ x-70 & \mbox{if } 70 \leq x \leq 100\\ \end{array}\right.$

Now, as we said before X represents the random variable where the bus breaks down, then we form a new random variable $Y = D(X)$ , $Y$ is a random variable as well, remember that there is a theorem that says that

$E[Y] = E[D(X)] = \int\limits_{-\infty}^{\infty} D(x) f(x) \,\, dx$

Where $f(x)$ is the probability mass function of X. Using the information of our problem

$E[Y] = \int\limits_{-\infty}^{\infty} D(x)f(x) dx \\= \frac{1}{100} \bigg[ \int\limits_{0}^{20} x dx +\int\limits_{20}^{45} (x-20) dx +\int\limits_{45}^{70} (70-x) dx +\int\limits_{70}^{100} (x-70) dx \bigg]\\= \frac{51}{4} = 12.75$

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3 years ago