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3 years ago

select all the faces of the right rectangle that have the same shape and dimensions as the cross section shown

Mathematics

2 answers:

LenKa [72]3 years ago

5 0

Answer:top,bottom

Step-by-step explanation:

Select all the faces of the right rectangular prism that have the same shape and dimensions as the cross section shown in the figure.

A rectangular prism with a vertical cross section drawn through the prism parallel to the sides.

Paul [167]3 years ago

3 0

Where’s the picture?

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What is the equivalent to 2:3 ?

leva [86]

2:3 is equivalent to 0.666 repeating

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3 years ago

suppose a charity received a donation of $22.8 million. if this represents 58% of the charity's donated fund, what is the total

Mrac [35]

The total amount is "x", so, if 22.8 is 58%, "x" is the 100%,

$\bf \begin{array}{ccllll}
amount&\%\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
22.8&58\\
x&100
\end{array}\implies \cfrac{22.8}{x}=\cfrac{58}{100}\implies \cfrac{22.8\cdot 100}{58}=x$

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3 years ago

Solve the problem for Part A and Part B

Novosadov [1.4K]

Part A answer is (A)

Part B answer is 3.170

5 0

3 years ago

What is -7.54 - 1.98 - 0.75

Vinil7 [7]

Answer:

-10.27

Step-by-step explanation:

-7.54-1.98=-9.52, and -9.52-0.75=-10.27

8 0

3 years ago

Read 2 more answers

A major cab company in Chicago has computed its mean fare from O'Hare Airport to the Drake Hotel to be $28.75 , with a standard

bulgar [2K]

Answer:

Following are the solution to the given choices:

Step-by-step explanation:

Using chebyshev's theorem :

$\to P(|(x-\mu)|\leq k \sigma)\geq 1- \frac{1}{k^2}\\\\here \\ \to \mu=28.75\\\\ \to \sigma=4.44$

In point a)

$\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}$

$= \frac{8.58}{4.44} \\\\ =1.9 \\\\ =2$

$value= (1- \frac{1}{k^2}) \times 100 \% =75\%$

In point b)

$\to (1- \frac{1}{k^2}) \times 100 \% =84\% \\\\\to (1- \frac{1}{k^2})=0.84 \\\\\to \frac{1}{k^2} =0.16 \\\\\to \frac{1}{k}=0.4\\\\ \to k=2.5 \\\\\to |(x-\mu)| \leq k \sigma \\\\= |(x-\mu)|\leq 2.5 \times 4.44 \\\\ = |(x-\mu)|\leq 1.11 \\\\ = (28.75\pm 11.1) \\\\\to \text{fares lies between}(17.65,39.85)$

In point c)

$\to 99.7 \% \\ lie \ between=28.75 \pm z(.03)\times \sigma \\\\ =28.75\pm 2.97 \times 4.44\\\\=(28.75\pm 13.1)\\\\=(15.65,41.85)\\$

In point d)

using standard normal variate

$x=20.17\\\\ z=-2 \\\\x=37.13\\\\ z=2\\\\\to P(20.17$

8 0

3 years ago