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Mazyrski [523]
4 years ago
9

(4,2);x=-3 how do you solve this problem

Mathematics
1 answer:
Alexandra [31]4 years ago
6 0
Find the equation of the line and plug in the x value that gives you which is -3? i am not very sure so i am sorry if i am wrong
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Where are the X intercept and Y intercept of the graph of y=1/3x-6
goldfiish [28.3K]

Answer:

x-int= (18,0), y-int= (0,-6)

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3 years ago
A ball is thrown upward from the top of a building. the function below shows the height of the ball in relation to sea level, ft
ser-zykov [4K]

average rate of change is the slope (of the line) between the two coordinates

f(t) = -16t² + 48t + 100

f(3) = -16(3)² + 48(3) + 100   = -144 + 144 + 100   = 100            (3, 100)

f(5) = -16(5)² + 48(5) + 100   = -400 + 240 + 100 = -60            (5, -60)

m = \frac{y2 - y1}{x2 - x1} = \frac{100 - (-60)}{3 - 5} = \frac{160}{-2} = -80

Answer: -80 ft/sec

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3 years ago
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Find the percentage of the markup. Round to the nearest tenth.<br><br> $33.00 to $45.00
babunello [35]

Answer:

0.7

Step-by-step explanation

Merry Christmas

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3 years ago
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The final scores of students in a graduate course are distributed normally with a mean of 72 and a standard deviation of 5. What
Tcecarenko [31]

Answer:

0.8041

Step-by-step explanation:

We know that  

μ=72  and  σ=5

and P(65<X<78)

We can determine the Z value as (X-μ)/σ

P( 65<X<78 )=P( 65-72< X-μ<78-72)

P((65-72)/5

P(65

To fine the Z values:

P (-1.4

From the standard normal tables:

P (Z

to find P ( Z<-1.4)

P ( Z

From the standard normal tables:

P ( Z

Therefore

P(-1.4

P (65

8 0
3 years ago
When rabbits were introduced to the continent of Australia they quickly multiplied and spread across the continent since there w
Lady bird [3.3K]

Answer:

a. y=6(1.7472)^x

b. y=6e^{0.558t}

c.13.3 months

Step-by-step explanation:

a.-Given the first term  at t_0 is 6 and the second term at t_3 is 32.

-Let's take rabbit population as a function of time to be

y=ab^x

where y is the population at time x and a the initial population at t_0\\

#We substitute our values to calculate the value of the constant b:

y_x=ab^x\\\\y_3=ab^3\\\\32=6b^3\\\\b=1.472

#Replace b in the population function:

y=ab^x, b=1.7472,a=6\\\\\therefore y=6(1.7472)^x

Hence, the regression for the rabbit population as a function of time x is y=6(1.7472)^x

b. The exponential function in terms of base e is usually expressed as:

A=A_0e^{kt}

Where:

A_0-is the initial population at t_o

A-is the population at time t.

k-is the  exponential growth constant.

e- the exponent

Our function in terms of base exponent is rewritten as:

y=A_0e^{kt}

#Substitute with actual figures to solve for t:

y=A_0e^{kt}, y=32, xt=3, A_0=6\\\\32=6e^{3k}\\\\3k=In (32/6)\\\\k=0.5580

Hence, the regression equation in terms of base e is y=6e^{0.558t}

c. We substitute y with any number higher than 10,000 to estimate the time for the rabbits to exceed 10,000.

-We know that y=6e^{0.558t}.

Therefore we calculate t as(take y=10001):

y=6e^{0.558t}, y=10001\\\\10001=6e^{0.558t}\\\\1666.8333=e^{0.558t}\\\\0.558t= In 1666.8333\\\\t=13.2951

Hence, it takes approximately 13.3 months for the population to exceed 10000

3 0
4 years ago
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