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2 years ago

Write the equation of the line in slope-intercept form.

Mathematics

1 answer:

Ymorist [56]2 years ago

8 0

Answer:

y = 1/40x + 20

Step-by-step explanation:

Slope-intercept equation format:

y = mx + b

m = slope/ gradient/ constant of variation/ rate of change

b = y-intercept

To find the slope, use the formula:

$\frac{y_2-y_1}{x_2-x_1}$

Find two random points:

1 - (0.5, 40) , 2 - (1, 60)

Plug in:

$\frac{60-40}{1-0.5}$

Solve:

$\frac{20}{0.5}$

Divide:

$\frac{1}{40}$

The y-intercept is when the x-coordinate is equal to 0.

The y-intercept is always the beginning.

So observing the graph, the first point is (0, 20)

The x-coordinate is 0.

Y intercept = 20.

Plug in both slope and y-intercept:

y = 1/40x + 20

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Answer:

A=3797 dollars

Step-by-step explanation:

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3 years ago

A sample of 300 urban adult residents of a particular state revealed 65 who favored increasing the highway speed limit from 55 t

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Answer and explanation:

Null hypothesis(H0) says sentiment for increasing speed limit in the two populations is the same

Alternative hypothesis(Ha) says sentiment for increasing speed limit in the two populations is different

Where p1 is the first population proportion =65/297= 0.22

And p2 is the second population proportion = 78/189= 0.41

P= p1+p2/n1+n2= 65+78/297+189= 0.29

Hence H0= p1-p2=0

Ha=p1-p2≠0

Test statistic= p1-p2/√p(1-p) (1/n1+1/n2)

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3 years ago

Kym plans to deposit $100 in an account at the end of each month for the next five years so that she can take a trip. (a) If Kym

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Answer:

Results are below.

Step-by-step explanation:

Giving the following information:

Monthly deposit= $100

Interest rate= 0.06/12= 0.005

Number of periods= 12*5= 60 months

a)

To calculate the future value, we need to use the following formula:

FV= {A*[(1+i)^n-1]}/i

A= monthly deposit

FV= {100*[(1.005^60) - 1]} / 0.005

FV= $6,977

b) If the deposit is at the beginning of the month, the interest is compounded one more period. We need to use the following formula:

FV= {A*[(1+i)^n-1]}/i + {[A*(1+i)^n]-A}

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3 years ago

A tank contains 60 kg of salt and 1000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain

MissTica

Answer:

(a) 60 kg; (b) 21.6 kg; (c) 0 kg/L

Step-by-step explanation:

(a) Initial amount of salt in tank

The tank initially contains 60 kg of salt.

(b) Amount of salt after 4.5 h

$\text{Let A = mass of salt after t min}\\\text{and }r_{i} = \text{rate of salt coming into tank}\\\text{and }r_{0} =\text{rate of salt going out of tank}$

(i) Set up an expression for the rate of change of salt concentration.

$\dfrac{\text{d}A}{\text{d}t} = r_{i} - r_{o}\\\\\text{The fresh water is entering with no salt, so}\\ r_{i} = 0\\r_{o} = \dfrac{\text{3 L}}{\text{1 min}} \times \dfrac {A\text{ kg}}{\text{1000 L}} =\dfrac{3A}{1000}\text{ kg/min}\\\\\dfrac{\text{d}A}{\text{d}t} = -0.003A \text{ kg/min}$

(ii) Integrate the expression

$\dfrac{\text{d}A}{\text{d}t} = -0.003A\\\\\dfrac{\text{d}A}{A} = -0.003\text{d}t\\\\\int \dfrac{\text{d}A}{A} = -\int 0.003\text{d}t\\\\\ln A = -0.003t + C$

(iii) Find the constant of integration

$\ln A = -0.003t + C\\\text{At t = 0, A = 60 kg/1000 L = 0.060 kg/L} \\\ln (0.060) = -0.003\times0 + C\\C = \ln(0.060)$

(iv) Solve for A as a function of time.

$\text{The integrated rate expression is}\\\ln A = -0.003t + \ln(0.060)\\\text{Solve for } A\\A = 0.060e^{-0.003t}$

(v) Calculate the amount of salt after 4.5 h

a. Convert hours to minutes

$\text{Time} = \text{4.5 h} \times \dfrac{\text{60 min}}{\text{1h}} = \text{270 min}$

b.Calculate the concentration

$A = 0.060e^{-0.003t} = 0.060e^{-0.003\times270} = 0.060e^{-0.81} = 0.060 \times 0.445 = \text{0.0267 kg/L}$

c. Calculate the volume

The tank has been filling at 6 L/min and draining at 3 L/min, so it is filling at a net rate of 3 L/min.

The volume added in 4.5 h is

$\text{Volume added} = \text{270 min} \times \dfrac{\text{3 L}}{\text{1 min}} = \text{810 L}$

Total volume in tank = 1000 L + 810 L = 1810 L

d. Calculate the mass of salt in the tank

$\text{Mass of salt in tank } = \text{1810 L} \times \dfrac{\text{0.0267 kg}}{\text{1 L}} = \textbf{21.6 kg}$

(c) Concentration at infinite time

$\text{As t $\longrightarrow \, -\infty,\, e^{-\infty} \longrightarrow \, 0$, so A $\longrightarrow \, 0$.}$

This makes sense, because the salt is continuously being flushed out by the fresh water coming in.

The graph below shows how the concentration of salt varies with time.

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