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3 years ago

Find the missing value for the exponential function represented by the table below x -2 -1 0 1 2 y 29 20.3 14.21 6.9629

Mathematics

1 answer:

andre [41]3 years ago

5 0

Answer:

y=4.8710 is the missing value

Step-by-step explanation:

The first step in approaching this question is determining the exponential equation that models the set of data. This can easily be done in Ms.Excel application. We first enter the data into any two adjacent columns of an excel workbook. The next step is to highlight the data, click on the insert tab and select the x,y scatter-plot feature. This creates a scatter-plot for the data.

The next step is to click the Add chart element feature and insert an exponential trend line to the scatter plot ensuring the display equation on chart is checked.

The exponential regression equation for the data set is given as;

$y=12.321e^{-0.464x}$

To find the missing y value, we simply substitute x with 2 in the regression equation obtained;

$y=12.321e^{-0.464(2)}\\\\y=4.8710$

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2/5x+1/4x=26 what is x

shepuryov [24]

Answer:

x = 40

Step-by-step explanation:

2/5x + 1/4x = 26

~Find common denominators

8/20x + 5/20x = 26

~Combine like terms

13/20x = 26

~Multiply 20/13 to both sides

x = 40

Best of Luck!

6 0

3 years ago

At 2:00 PM a car's speedometer reads 30 mi/h. At 2:20 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:20 the acce

Bad White [126]

Answer:

Let v(t) be the velocity of the car t hours after 2:00 PM. Then $\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }$ . By the Mean Value Theorem, there is a number c such that $0 < c$ with $v'(c)=60 \:{\frac{mi}{h^2}}$ . Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly $60 \:{\frac{mi}{h^2}}$ .

Step-by-step explanation:

The Mean Value Theorem says,

Let be a function that satisfies the following hypotheses:

f is continuous on the closed interval [a, b].
f is differentiable on the open interval (a, b).

Then there is a number c in (a, b) such that

$f'(c)=\frac{f(b)-f(a)}{b-a}$

Note that the Mean Value Theorem doesn’t tell us what c is. It only tells us that there is at least one number c that will satisfy the conclusion of the theorem.

By assumption, the car’s speed is continuous and differentiable everywhere. This means we can apply the Mean Value Theorem.

Let v(t) be the velocity of the car t hours after 2:00 PM. Then $v(0 \:h) = 30 \:{\frac{mi}{h} }$ and $v( \frac{1}{3} \:h) = 50 \:{\frac{mi}{h} }$ (note that 20 minutes is $20/60=1/3$ of an hour), so the average rate of change of v on the interval $[0 \:h, \frac{1}{3} \:h]$ is

$\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }$

We know that acceleration is the derivative of speed. So, by the Mean Value Theorem, there is a time c in $(0 \:h, \frac{1}{3} \:h)$ at which $v'(c)=60 \:{\frac{mi}{h^2}}$ .