All categories

1 year ago

Find the rate of change for the situation. You run 14 miles in two hours and 28 miles in four hours.

Mathematics

1 answer:

krek1111 [17]1 year ago

4 0

the slope goes by several names

• average rate of change

• rate of change

• deltaY over deltaX

• Δy over Δx

• rise over run

• gradient

• constant of proportionality

however, is the same cat wearing different costumes.

$\begin{array}{ccll} \stackrel{x}{miles}&\stackrel{y}{hours}\\ \cline{1-2} 14&2\\ 28&4 \end{array}\qquad \qquad \begin{array}{llll} (\stackrel{x_1}{14}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{28}~,~\stackrel{y_2}{4}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{4}-\stackrel{y1}{2}}}{\underset{run} {\underset{x_2}{28}-\underset{x_1}{14}}}\implies \cfrac{2}{14}\implies \cfrac{1}{7} \end{array}$

You might be interested in

Tom spent 153 minutes completing a race. He walked 63 minutes and jogged the rest. What is the ratio of time he jogged to time h

ra1l [238]

Subtract 153-63=90 and I am pretty sure that's what you have to do

5 0

3 years ago

Read 2 more answers

Find b. Question 19 options: 70.1 57.7 31.5 43.8

Reptile [31]

Answer: 57.7

Step-by-step explanation:

6 0

3 years ago

The port of South Louisiana, located along 54 miles of the Mississippi River between New Orleans and Baton Rouge, is the largest

Ksenya-84 [330]

Answer:

a) 0.7287

b) 0.9663

c) 0.237

d) 3.65 tons of cargo per week or more that will require the port to extend its operating hours.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4.5 million tons of cargo per week

Standard Deviation, σ = 0 .82 million tons

We are given that the distribution of number of tons of cargo handled per week is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P( port handles less than 5 million tons of cargo per week)

P(x < 5)

$P( x < 5) = P( z < \displaystyle\frac{5 - 4.5}{0.82}) = P(z < 0.609)$

Calculation the value from standard normal z table, we have,

$P(x < 5) =0.7287= 72.87\%$

b) P( port handles 3 or more million tons of cargo per week)

$P(x \geq 3) = P(z \geq \displaystyle\frac{3-4.5}{0.82}) = P(z \geq −1.82926)\\\\P( z \geq −1.82926) = 1 - P(z < -1.829)$

Calculating the value from the standard normal table we have,

$1 - 0.0337 = 0.9663 = 96.63\%\\P( x \geq 3) = 96.63\%$

c)P( port handles between 3 million and 4 million tons of cargo per week)

$P(3 \leq x \leq 4) = P(\displaystyle\frac{3 - 4.5}{0.82} \leq z \leq \displaystyle\frac{4-4.5}{0.82}) = P(-1.829 \leq z \leq -0.609)\\\\= P(z \leq -0.609) - P(z < -1.829)\\= 0.271-0.034 = 0.237= 23.7\%$