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3 years ago

True or false: the height is always twice the length of the base edge for any triangular pyramid

Mathematics

1 answer:

AleksandrR [38]3 years ago

8 0

Answer:

FALSE

Step-by-step explanation:

The length of the base edge of a triangular pyramid may vary.

The height is not always twice the length of the base edge for any triangular pyramid, the dimensions are not fixed for any shape and can vary from one design to another.

more so, i may require a particular size or dimension of triangular based pyramid for my application that may not suit your application.

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Answer:

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Step-by-step explanation:

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Work out the value of 10 fourths

lora16 [44]

Answer:

10/4 equals 2 1/2 when simplified.

Step-by-step explanation:

10/4 = 2 2/4 or 2 1/2 when simplified.

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2 years ago

Find and interpret the mean absolute deviation of the data. round your answer to the nearest tenth if necessary.

natka813 [3]

Answer:

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Step-by-step explanation:

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2 years ago

Let X be the number of the cars being repaired at a repair shop. We have the following information:

worty [1.4K]

Answer:

(a) Sample Space

$S = \{0,1,2,3\}$

(b) PMF

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {P(x)} & {1/3} & {1/6} & {1/6} & {1/3} \ \end{array}$

(c) CDF

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {F(x)} & {1/3} & {1/2} & {2/3} & {1} \ \end{array}$

Step-by-step explanation:

Solving (a): The sample space

From the question, we understand that at most 3 cars will be repaired.

This implies that, the number of cars will be 0, 1, 2 or 3

So, the sample space is:

$S = \{0,1,2,3\}$

Solving (b): The PMF

From the question, we have:

$P(2) = P(1)$

$P(0) = P(3)$

$P(1\ or\ 2) = 0.5 * P(0\ or\ 3)$

$P(1\ or\ 2) = 0.5 * P(0\ or\ 3)$ can be represented as:

$P(1) + P(2) = 0.5[P(0) + P(3)]$

Substitute $P(2) = P(1)$ and $P(0) = P(3)$

$P(1) + P(1) = 0.5[P(0) + P(0)]$

$2P(1) = 0.5[2P(0)]$

$2P(1) = P(0)$

$P(0)= 2P(1)$

Also note that:

$P(0) + P(1) + P(2) + P(3) = 1$

Substitute $P(2) = P(1)$ and $P(0) = P(3)$

$P(0) + P(1) + P(1) + P(0) = 1$

$2P(1) + 2P(0) = 1$

Substitute $P(0)= 2P(1)$

$2P(1) + 2*2P(1) = 1$

$2P(1) + 4P(1) = 1$

$6P(1) = 1$

Solve for P(1)

$P(1) = \frac{1}{6}$

To calculate others, we have:

$P(2) = P(1)$

$P(2) = P(1) = \frac{1}{6}$

$P(0)= 2P(1)$

$P(0) =2 * \frac{1}{6}$ $P(0) =\frac{1}{3}$

$P(3) = P(0) =\frac{1}{3}$

Hence, the PMF is:

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {P(x)} & {1/3} & {1/6} & {1/6} & {1/3} \ \end{array}$

<em>See attachment (1) for histogram</em>

Solving (c): The CDF ; F(x)

This is calculated as:

$F(x) = P(X \le x) =\sum\limit^{3}_{x_i \le x} P(x_i)$

For x = 0;

We have:

$P(X \le 0) = P(0)$

$P(X \le 0) = 1/3$

For x = 1

$P(X \le 1) = P(0) + P(1)$

$P(X \le 1) = 1/3 + 1/6$

$P(X \le 1) = 1/2$

For x = 2

$P(X \le 2) = P(0) + P(1) + P(2)$

$P(X \le 2) = 1/3 + 1/6 + 1/6$

$P(X \le 2) = 2/3$

For x = 3

$P(X \le 3) = P(0) + P(1) + P(2) + P(3)$

$P(X \le 3) = 1/3 + 1/6 + 1/6 + 1/3$

$P(X \le 3) = 1$

Hence, the CDF is:

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {F(x)} & {1/3} & {1/2} & {2/3} & {1} \ \end{array}$

<em>See attachment (2) for histogram</em>

6 0

2 years ago