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Deffense [45]
3 years ago
11

jake is hiking a trail that leads to the top of a canyon. The trail 4.2 miles long, and jake plans to stop for lunch after he co

mpletes 1.6 miles. How far from the top of the canyon will jake be when he stops for lunch?
Mathematics
1 answer:
Ksivusya [100]3 years ago
5 0
The answer will be 2.2 miles

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Twenty-six , 2 tens 6 ones , 25, 20+6 which one is the odd one out?
Aneli [31]

Answer:

25

Step-by-step explanation:

twenty-six=26

2 tens 6 ones= 26

25=25

20+6=26

25 is the odd one out.

8 0
3 years ago
What is .5 as a fraction
NNADVOKAT [17]

Answer:

1/2 is 0.5 as a fraction


8 0
3 years ago
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There are 6 roses in a vase of 11 flowers. The rest are daisies. What is the ratio of all flowers in the vase to daisies (PICTUR
lara [203]

Answer:

a) 6:5

b) 11:5

(Daisies: 11-6=5)

5 0
3 years ago
What is the value of the product (3-2i)(3+2i)?
Helga [31]

Answer:

13

Step-by-step explanation:

(3-2i)(3+2i) = 9  + 6i - 6i - 4i^2

9 - 4i^2

9 + 4 = 13

5 0
4 years ago
Read 2 more answers
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
4 years ago
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