What is the exact volume of the cylinder?, (a) 54in. (b) 108in. (c) 162in. (d) 324in.

A coin will be tossed three times, and each toss will be recorded as heads (H) or tails (7). Give the sample space describing al

Fynjy0 [20]

Answer:

A). {HHH, HHT, HTH, THH, TTH,THT,HTT,TTT}

(B). {HHH, HHT, THH, THT}

Step-by-step explanation:

Given:

A coin is to be tossed three times and each tossed will be recorded as Head(H) and Tails(T).

To Find:

(a). Sample Space

(b). Event that the second toss is Heads

Explanation:

(A). Sample Space: When a coin is tossed 3 times

{HHH, HHT, HTH, THH, TTH,THT,HTT,TTT}

(B). Event that second toss is Head

{HHH, HHT, THH, THT}

6 0

3 years ago

Let X denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose t

andrezito [222]

Answer and Step-by-step explanation: For an exponential distribution, the probability distribution function is:

f(x) = λ. $e^{-\lambda.x}$

and the cumulative distribution function, which describes the probability distribution of a random variable X, is:

F(x) = 1 - $e^{-\lambda.x}$

(a) Probability of distance at most 100m, with λ = 0.0143:

F(100) = 1 - $e^{-0.0143.100}$

F(100) = 0.76

Probability of distance at most 200:

F(200) = 1 - $e^{-0.0143.200}$

F(200) = 0.94

Probability of distance between 100 and 200:

F(100≤X≤200) = F(200) - F(100)

F(100≤X≤200) = 0.94 - 0.76

F(100≤X≤200) = 0.18

(b) The mean, E(X), of a probability distribution is calculated by:

E(X) = $\frac{1}{\lambda}$

E(X) = $\frac{1}{0.0143}$

E(X) = 69.93

The standard deviation is the square root of variance,V(X), which is calculated by:

σ = $\sqrt{\frac{1}{\lambda^{2}} }$

σ = $\sqrt{\frac{1}{0.0143^{2}} }$

σ = 69.93

Distance exceeds the mean distance by more than 2σ:

P(X > 69.93+2.69.93) = P(X > 209.79)

P(X > 209.79) = 1 - P(X≤209.79)

P(X > 209.79) = 1 - F(209.79)

P(X > 209.79) = 1 - (1 - $e^{-0.0143*209.79}$ )

P(X > 209.79) = 0.0503

(c) Median is a point that divides the value in half. For a probability distribution:

P(X≤m) = 0.5

$\int\limits^m_0 f({x}) \, dx$ = 0.5

$\int\limits^m_0 {\lambda.e^{-\lambda.x}} \, dx$ = 0.5

$\lambda.\frac{e^{-\lambda.x}}{-\lambda}$ = $-e^{-\lambda.x} + e^{0}$

$1 - e^{-\lambda.m}$ = 0.5

$-e^{-\lambda.m}$ = - 0.5

ln( $e^{-0.0143.m}$ ) = ln(0.5)

-0.0143.m = - 0.0693

m = 48.46

6 0

3 years ago

1) If the designer measures the pond and determines that the length of the pond across is 400

ozzi

Answer: 200

Step-by-step explanation: to get the radius you divided by 2. bc there is only 2 sides , have a great day.

3 0

2 years ago

For pigs, the length of pregnancies varies according to a normal distribution with a mean of 114 days and a standard deviation o

riadik2000 [5.3K]

Answer: The length of approximately 68% of all pig pregnancies will fall between 109 days and 119 days.

Step-by-step explanation:

According to the empirical rule , 68% of the population falls within one standard deviations from the mean.

Given : For pigs, the length of pregnancies varies according to a normal distribution with a mean of 114 days and a standard deviation of 5 days.

According to the Empirical Rule, the length of approximately 68% of all pig pregnancies will fall between $114-1(5)$ days and $114+1(5)$ days .

i.e. the length of approximately 68% of all pig pregnancies will fall between 109 days and 119 days.

7 0

2 years ago

Help on math plzzzz....

NARA [144]

5 is c
6 is d
7 is a
Ghfxghxfgdgsfgshsdghfsghfghsfgsgfh

8 0

2 years ago