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3 years ago

A designer creates a model of a building. The model is 63.5 centimeters long. One inch equals 2.54 centimeters.

Mathematics

1 answer:

Yakvenalex [24]3 years ago

8 0

Answer:

25in

Step-by-step explanation:

if the model is 63. 5 cm, and 2.54 cm is equivalent to 1 in

you can do 63.5 / 2.54 to get 25 in

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sesenic [268]

Answer:

C is the correct answer cos it is paid by the customer

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3 years ago

Identify whether the given pairs of events in a roulette game are mutually exclusive. Drag the items on the left to the correct

xxTIMURxx [149]

Answer:

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not mutually exclusive events 1 to 12 and red

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mutually exclusive events 1 to 12 and 25 to 36

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mutually exclusive events black and red

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not mutually exclusive events even and black

Step-by-step explanation:

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3 years ago

The question is on the attached image :)

olga_2 [115]

The mean can be found by adding the two numbers and dividing by 2

let the second number be y

(x + y)/2 = 1/2x + 1

solve for y

multiply each side by 2
x + y = 2(1/2x + 1)
distribute
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ANSWER: the second number is 2

6 0

3 years ago

Suppose we roll a fair die and let X represent the number on the die. (a) Find the moment generating function of X. (b) Use the

Likurg_2 [28]

Answer:

(a) moment generating function for X is $\frac{1}{6}\left(e^{t}+e^{2 t}+e^{2 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$

(b) $\mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}$

Step-by step explanation:

Given X represents the number on die.

The possible outcomes of X are 1, 2, 3, 4, 5, 6.

For a fair die, $P(X)=\frac{1}{6}$

(a) Moment generating function can be written as $M_{x}(t)$ .

$M_x(t)=\sum_{x=1}^{6} P(X=x)$

$M_{x}(t)=\frac{1}{6} e^{t}+\frac{1}{6} e^{2 t}+\frac{1}{6} e^{3 t}+\frac{1}{6} e^{4 t}+\frac{1}{6} e^{5 t}+\frac{1}{6} e^{6 t}$

$M_x(t)=\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$

(b) Now, find $E(X) \text { and } E\((X^{2})$ using moment generating function

$M^{\prime}(t)=\frac{1}{6}\left(e^{t}+2 e^{2 t}+3 e^{3 t}+4 e^{4 t}+5 e^{5 t}+6 e^{6 t}\right)$

$M^{\prime}(0)=E(X)=\frac{1}{6}(1+2+3+4+5+6)$

$\Rightarrow E(X)=\frac{21}{6}$

$M^{\prime \prime}(t)=\frac{1}{6}\left(e^{t}+4 e^{2 t}+9 e^{3 t}+16 e^{4 t}+25 e^{5 t}+36 e^{6 t}\right)$

$M^{\prime \prime}(0)=E(X)=\frac{1}{6}(1+4+9+16+25+36)$

$\Rightarrow E\left(X^{2}\right)=\frac{91}{6}$

Hence, (a) moment generating function for X is $\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$ .

(b) $\mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}$

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4 years ago

The values shown represent the y-values that correspond to x-values from x = 1 through x = 5.

rjkz [21]

The answer is B because every number is adding up to 6

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3 years ago