What dimensions will produce a square-based box, of minimum surface area, with volume of

Basic Computation: Find Probabilities In Problems 5-14, assume that x has a normal distribution with the specified mean and stan

Ulleksa [173]

Answer:

the answer is below

Step-by-step explanation:

The z score is used to calculate by how many standard deviations the raw score is above or below the mean. The z score is given as:

$z=\frac{x-\mu}{\sigma}\\\\\mu=mean,\sigma=standard\ deviation$

1) For x = 3

$z=\frac{x-\mu}{\sigma}=\frac{3-4}{2}=-0.5$

For x = 6

$z=\frac{x-\mu}{\sigma}=\frac{6-4}{2}=1$

P(3 ≤ x ≤ 6) = P(-0.5 ≤ z ≤ 1) = P(z < 1) - P(z < -0.5) = 0.8413 - 0.3085 = 0.5328

2) For x = 50

$z=\frac{x-\mu}{\sigma}=\frac{50-40}{15}=0.67$

For x = 70

$z=\frac{x-\mu}{\sigma}=\frac{70-40}{15}=2$

P(50 ≤ x ≤ 70) = P(0.67 ≤ z ≤ 2) = P(z < 2) - P(z < 0.67) = 0.9772 - 0.7486 = 0.2286

3) For x = 8

$z=\frac{x-\mu}{\sigma}=\frac{8-15}{3.2}=-2.19$

For x = 12

$z=\frac{x-\mu}{\sigma}=\frac{12-15}{3.2}=-0.94$

P(8 ≤ x ≤ 12) = P(-2.19 ≤ z ≤ -0.94) = P(z < -0.94) - P(z < -2.19) = 0.1736 - 0.0143 = 0.1593

4) For x = 30

$z=\frac{x-\mu}{\sigma}=\frac{30-20}{3.4}=2.94$

P(x ≥ 30) = P(z ≥ 2.94) = 1 - P(z < 2.94) = 1 - 0.9984 = 0.0016

5) x = 90

$z=\frac{x-\mu}{\sigma}=\frac{90-100}{15}=-0.67$

P(x ≥ 90) = P(z ≥ -0.67) = 1 - P(z < -0.67) = 1 - 0.2514 = 0.7486

6) For x = 10

$z=\frac{x-\mu}{\sigma}=\frac{10-15}{4}=-1.25$

For x = 20

$z=\frac{x-\mu}{\sigma}=\frac{20-15}{4}=1.25$

P(10 ≤ x ≤ 20) = P(-1.25 ≤ z ≤ 1.25) = P(z < 1.25) - P(z < -1.25) = 0.8944 - 0.1056 = 0.7888

7) For x = 7

$z=\frac{x-\mu}{\sigma}=\frac{7-5}{1.2}=1.67$

For x = 9

$z=\frac{x-\mu}{\sigma}=\frac{9-5}{1.2}=3.33$

P(7 ≤ x ≤ 9) = P(1.67 ≤ z ≤ 3.33) = P(z < 3.33) - P(z < 1.67) = 0.9996 - 0.9525 = 0.0471

8) For x = 40

$z=\frac{x-\mu}{\sigma}=\frac{40-50}{15}=-0.67$

For x = 47

$z=\frac{x-\mu}{\sigma}=\frac{47-50}{15}=-0.2$

P(40 ≤ x ≤ 47) = P(-0.67 ≤ z ≤ -0.2) = P(z < -0.2) - P(z < -0.67) = 0.4207 - 0.2514 = 0.1693

9) x = 120

$z=\frac{x-\mu}{\sigma}=\frac{120-10}{15}=7.33$

P(x ≥ 120) = P(z ≥ 7.33) = 1 - P(z < 7.33) = 1 - 0.9999 = 0.001

10) x = 2

$z=\frac{x-\mu}{\sigma}=\frac{2-3}{0.25}=-4$

P(x ≥ 2) = P(z ≥ -4) = 1 - P(z < -4) = 1 - 0.0001 = 0.999

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A marine biologist is studying the growth of a particular species of fish

: When the marine biologist concluded her study, the length of the fish was approximately 9.19 cm

To Find : reasonable domain to plot the growth function

What does the y-intercept of the graph of the function f(m) represent?

average rate of change of the function f(m) from m = 1 to m = 9,

Solution:

m = 0

=> f(0) = 5(1.07)⁰ = 5 cm

the length of the fish was approximately 9.19 cm.

=>

=> m = 9

Domain = [0 , 9 ]

reasonable domain to plot the growth function = [0 , 9 ]

y-intercept of the graph of the function f(m) represent represent initial length of fish = 5 cm

average rate of change of the function f(m) from m = 1 to m = 9

m = 1 => f(1) = 5(1.07) = 5.35 cm

m = 9 => f(9) = 5(1.07)⁹ = 9.19 cm

Change in 9 - 1 = 8 months = 9.19 - 5.35 = 3.84 cm

average rate of change of the function f(m) from m = 1 to m = 9,

= (3.84/8)

= 0.48 cm per month

Step-by-step explanation:

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