Using the formula
A = Pe^kt
Where A is the ending amount of population, P is the population k is a constant and 't' is time.

A = 0.39, t= 12 and p= 0.29
0.39= 0.29e^12k
1.34 = e^12k
ln1.34 = 12k
ln1.34/12 = k
K= 0.02

No the graph does not support his claim as k should be equal to 0.5

4 0

3 years ago

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 6.9 years, and standard deviation of

Lostsunrise [7]

ANSWER

0.9738 or 97.38%

EXPLANATION

Given:

$\begin{gathered} mean(\mu)=6.9 \\ Standard\text{ }Deviation(\sigma)=1.6 \end{gathered}$

Desired Outcome:

Probability that it will last 10 years

z-score for the sample:

$\begin{gathered} z-score=\frac{X-\mu}{\sigma} \\ z-score=\frac{10-6.9}{1.6} \\ z-score=1.9376 \end{gathered}$

p-value

For the z-score of 1.9376, the p-value is 0.9738 or 97.38%

Hence, the probability that it will last longer than 10 years if you randonly purchase one item is 97.38%

7 0

1 year ago

Solve the equation below for the value of x.<br><br> −314x + 25 = 214x − 1312

gogolik [260]

Answer:

x=1337/528

Step-by-step explanation:

-314x-214x=-1312-25

-528x=-1337

x=1337/528

8 0

3 years ago

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