Write the prime factorization of 315.

Suppose X has an exponential distribution with mean equal to 23. Determine the following:

e-lub [12.9K]

Answer:

a) P(X > 10) = 0.6473

b) P(X > 20) = 0.4190

c) P(X < 30) = 0.7288

d) x = 68.87

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

Mean equal to 23.

This means that $m = 23, \mu = \frac{1}{23} = 0.0435$

(a) P(X >10)

$P(X > 10) = e^{-0.0435*10} = 0.6473$

So

P(X > 10) = 0.6473

(b) P(X >20)

$P(X > 20) = e^{-0.0435*20} = 0.4190$

So

P(X > 20) = 0.4190

(c) P(X <30)

$P(X \leq 30) = 1 - e^{-0.0435*30} = 0.7288$

So

P(X < 30) = 0.7288

(d) Find the value of x such that P(X > x) = 0.05

So

$P(X > x) = e^{-\mu x}$

$0.05 = e^{-0.0435x}$

$\ln{e^{-0.0435x}} = \ln{0.05}$

$-0.0435x = \ln{0.05}$

$x = -\frac{\ln{0.05}}{0.0435}$

$x = 68.87$

5 0

3 years ago

100 POINTSSS! ASAP ANSWEERR PLS

Margarita [4]

PART A

Given:

f(x) = 0.69(1.03)x

To find:

If the price of the product is increasing or decreasing and by what percentage

Steps:

we know the formula to find the price of Product A per year, so

f(1) = 0.69 * 1.03 * 1

Price = $0.7107

f(2) = 0.69 * 1.03 * 2

Price = $1.4214

Here the Price of Product after 2 years is greater than the price of Product after one year. So the price of the product A is increasing.

Now to find percentage increase,

Percentage increase = $\frac{FV-SV}{SV}*100$ (FV = final value, SV = starting value)

Percentage increase = $\frac{1.4214 - 0.7107}{0.7107}*100$

Percentage increase = $\frac{0.7107}{0.7107}*100$

Percentage increase = 100 %

Therefore, the percentage increase of Product A is 100%

PART B

Given:

Price of product B in 1st year = $10,100

Price of product B in 2nd year = $10,201

Price of product B in 3rd year = $10,303.01

Price of product B in 4th year = $10,406.04

To find:

Which product recorded a greater percentage change over the previous year

Steps:

We need to find the percentage change of Product B and Product A of each year. We know that the percentage change of product A is 100 % for each year, so we only need to calculate for product B

PC of product B from 1st to 2nd year = $\frac{10,201-10,100}{10,100}*100$

= $\frac{101}{10,100}*100$

= 0.01 * 100

= 1 %

PC of product B from 2nd to 3rd year = $\frac{10,303.01-10,201}{10,201} *100$

= 1%

PC of product B from 3rd to 4th year $=\frac{10,406.04-10,303.01}{10,303.01}*100$

≈ 1%

So, percentage change of product B is 1% per year

Therefore, Product A has greater percentage change

Happy to help :)

If u need more help, feel free to ask

6 0

3 years ago

The Mean Corporation's statisticians would like to construct a hypothesis test for the mean annual profit (μ) earned by health-d

Jobisdone [24]

Missing Part of Question

Juice that is opened. It is known that the annual profits earned by health-drink franchises has a population standard deviation of $8,700

........

Answer:

z = 1.157

Step-by-step explanation:

Given

Annual profit as calculated = $90,300.

H0: μ = 89,000

Ha: μ > 89,000

σ = 8700

n = 60

Calculate the test statistic (z) that corresponds to the sample and hypotheses.

Test statistic (z) is calculated by

z = (x - μ)/(σ/√n)

Where x = 90,300

μ = 89,000

σ = 8700

n = 60

z = (90,300 - 89,000)/(8700/√60)

z = 1.157443298866584

z = 1.157 ----- Approximated

7 0

3 years ago

Question 1 of 10

yulyashka [42]

the answer is C

C. (4x + 5)(4x + 5)

5 0

2 years ago

Read 2 more answers

The Frostburg-Truth bus travels on a straight road from Frostburg Mall to Sojourner Truth Park. The mall is 3 miles west and 4 m

geniusboy [140]

We take the distance from points which indicates the location of the park and the mall.

For distance through north and east, we have positive values and negative for west and south.

Mall: (-3, -4)
Park: (3, 5)

The distance is calculate through the equation,
d = sqrt ((x₂ - x₁)² + (y₂ - y₁)²)

Substituting,
d = sqrt ((-4 - 5)² + (-3 - 3)²
d = sqrt 117 = 10.82
Thus, the distance between the mall and the park is approximately 10.82 miles.

5 0

3 years ago