The y intercept of the graph of the exponential decay equation

What customary unit would you use to estimate? A car is about 15 _____ long. inches meters yards feet

ch4aika [34]

Answer:

feet

Step-by-step explanation:

3 0

3 years ago

In a mapping diagram each _____ corresponds to an ordered pair. A.) input value B.)relation C.)arrow D.)output value

Lilit [14]

The answer is c. relation

5 0

3 years ago

Jack invested some money in a bank at a fixed rate of interest compounded annually. The equation below shows the value of his in

schepotkina [342]

Answer: The average rate of change of Jack's investment from the third year to the fifth year is $6.43

Step-by-step explanation:

The function that defines the value of his investment after x years,

Putting the value of x as 3 and 5, we can get the value of his investment after 3 years and 5 years respectively.

6 0

2 years ago

I need help with my geometry work please.

cricket20 [7]

Answer:

AA

Step-by-step explanation:

The steps are F and H are matched with line marking it congruent and G is vertical and the last is congruent.

5 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