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faltersainse [42]

3 years ago

11

A triangular prism has a triangle Base with a base of 35cm and a height 24cm. The prism has a height of 80cm. What is the Volume

of the prism?
V= cm
what is the volume

1 answer:

Agata [3.3K]3 years ago

3 0

67,200cm I took the test :)

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PtichkaEL [24]

It's two. 50% chance of heads, 50% chance of tails.

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2 years ago

Sawarasenikimiwa⛓shoujonano?✨böKùWâ‍♀️ÿARiçHiñBįCChĪńOoSû‍♀️DàYO

LiRa [457]

Step-by-step explanation:

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3 years ago

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Solve the following equation for the variable x. ( 1/27 is a fraction )

igor_vitrenko [27]

The answer is,
$\frac{10935}{59048}$
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3 years ago

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

6 0

3 years ago

The marketing club at school is opening a student store. They randomly survey 50 students about how much money they spend on lun

Luden [163]

Answer:

The expected value for a student to spend on lunch each day = $5.18

Step-by-step explanation:

Given the data:

Number of students______$ spent

2 students______________$10

1 student________________$8

12 students______________$6

23 students______________$5

8 students_______________$4

4 students_______________$3

Sample size, n = 50.

Let's first find the value on each amount spent with the formula:

$\frac{num. of students}{sample size} * dollar spent$

Therefore,

For $10:

$\frac{2}{50} * 10 = 0.4$

For $8:

$\frac{1}{50} * 8 = 0.16$ l

For $6:

$\frac{12}{50} * 6 = 1.44$

For $5:

$\frac{23}{50} * 5 = 2.3$

For $4:

$\frac{8}{50} * 4 = 0.64$

For $3:

$\frac{4}{50} * 3 = 0.24$

To find the expected value a student spends on lunch each day, let's add all the values together.

Expected value =

$0.4 + $0.16 + 1.44 +$2.3 + $0.64 + $0.24

= $5.18

Therefore, the expected value for a student to spend on lunch each day is $5.18

7 0

2 years ago