if it rains tomorrow, the probability is 0.8 that john will practice the piano. if it does not rain tomorrow, there is only a .4 chance that john will practice. if there is a 60% that it will rain tomorrow, what is the probability that John will practice his piano lesson? i'm supposed to use a tree diagram to solve this ...?

The value of k in the given function is k =3

From the question, the given function is F(x)= kx2

This can be properly written as

[tex]f(x) =kx^{2}[/tex]

Also, from the question, we have that f(2) = 12

Since [tex]f(x) =kx^{2}[/tex]

∴ [tex]f(2) =k(2)^{2}[/tex]

This becomes

[tex]f(2) = k \times 4[/tex]

[tex]f(2) = 4k[/tex]

Now, to determine the value of k, we will input the value of f(2), that is f(2)=12 in the above equation, that is

[tex]f(2) = 4k[/tex] becomes

[tex]12= 4k[/tex]

Now, divide both sides by 4

[tex]\frac{12}{4} = \frac{4k}{4}[/tex]

[tex]3 = k[/tex]

∴ [tex]k = 3[/tex]

Hence, the value of k in the given function is k =3

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Answer:

No, it is 3x-5 and r=0

Step-by-step explanation:

We can do it by long division method  the required quotient is 3x-5 and r=0

not 3x-1 , r=1

multiply the divisor with 3x  we will get [tex]3x^2+9x[/tex]to cancel out the first term of dividend

Now after solving we will get [tex]-5x-15[/tex]

Now, multiply the divisor by -5 we will get -5x-15 which will cancel the entire dividend.

Answer:

the answer is a hope it helps.

Step-by-step explanation:

Answer: 36.1 hours

Step-by-step explanation:

Given: The amount someone earns for each hour worked = [tex]\$6.25[/tex]

The expected amount to earn by work = [tex]\$225.65[/tex]

Now, to find the number of hours work to earn the expected value , we divide the expected value by the hourly rate, we get

The  number of hours work to earn [tex]\$225.65\ =\frac{225.65}{6.25}=36.104\approx36.1[/tex]  

Hence, the number of hours work to earn  [tex]\$225.65[/tex] about 36.1 hours.

The rate at which the epidemic is growing after 10 years is approximately 0.79.

Using the provided formula for the derivative of the population function with respect to time and evaluating it at ( t = 10), we have:

[tex]\[ \frac{dP}{dt} \Bigg|_{t=10} = \frac{-500(20,000)(-0.549)e^{-5.49}}{(1 + 20,000e^{-5.49})^2} \][/tex]

[tex]\[ \approx \frac{-500(20,000)(-0.549)e^{-5.49}}{(1 + 20,000e^{-5.49})^2} \][/tex]

[tex]\[ \approx \frac{-500(20,000)(-0.549)(0.004088)}{(1 + 20,000(0.004088))^2} \][/tex]

[tex]\[ \approx \frac{-500(20,000)(-0.549)(0.004088)}{(1 + 81.76)^2} \][/tex]

[tex]\[ \approx \frac{-500(20,000)(-0.549)(0.004088)}{(82.76)^2} \][/tex]

[tex]\[ \approx \frac{-500(20,000)(-0.549)(0.004088)}{6856.8976} \][/tex]

[tex]\[ \approx \frac{5431.56}{6856.8976} \][/tex]

[tex]\[ \approx 0.7926 \][/tex]

Rounding to two significant digits, the rate at which the epidemic is growing after 10 years is approximately 0.79.

Answer:

To find the rate of growth of the epidemic after 10 years, we'll first differentiate the epidemic curve equation with respect to time (t) and then plug in t = 10 to find the growth rate.

Therefore, after 10 years, the epidemic is growing at a rate of approximately -0.082 (rounded to two significant digits).

Step-by-step explanation:

To determine the rate of growth of the epidemic after 10 years, we'll first differentiate the given epidemic curve equation with respect to time (t) using the quotient rule and the chain rule of differentiation.

Let [tex]\( P = \frac{500}{1 + 20,000e^{-0.549t}} \)[/tex].

To differentiate P with respect to t, we'll use the quotient rule:

[tex]\[ \frac{dP}{dt} = \frac{d}{dt} \left( \frac{500}{1 + 20,000e^{-0.549t}} \right) \]\[ = \frac{0 - 500 \times \frac{d}{dt}(1 + 20,000e^{-0.549t})}{(1 + 20,000e^{-0.549t})^2} \][/tex]

Now, we'll find [tex]\( \frac{d}{dt}(1 + 20,000e^{-0.549t}) \)[/tex] using the chain rule:

[tex]\[ \frac{d}{dt}(1 + 20,000e^{-0.549t}) = 0 - 20,000 \times (-0.549)e^{-0.549t} \]\[ = 10,980e^{-0.549t} \][/tex]

Substituting this back into the differentiation of P:

[tex]\[ \frac{dP}{dt} = \frac{-500 \times 10,980e^{-0.549t}}{(1 + 20,000e^{-0.549t})^2} \][/tex]

Now, we'll find the growth rate after 10 years by plugging in [tex]\( t = 10 \)[/tex] into [tex]\( \frac{dP}{dt} \)[/tex]:

[tex]\[ \frac{dP}{dt} \bigg|_{t=10} = \frac{-500 \times 10,980e^{-0.549 \times 10}}{(1 + 20,000e^{-0.549 \times 10})^2} \]\[ \approx \frac{-500 \times 10,980 \times e^{-5.49}}{(1 + 20,000e^{-5.49})^2} \]\[ \approx \frac{-500 \times 10,980 \times 0.004056}{(1 + 20,000 \times 0.004056)^2} \]\[ \approx -0.082 \][/tex]

Thus, after 10 years, the epidemic is growing at a rate of approximately -0.082 (rounded to two significant digits).

By using the distance formula, the length of BB' on the graph is equal to [tex]\sqrt{29}\;units.[/tex]

In Mathematics and Geometry, the distance between two (2) end points that are on a coordinate plane can be calculated by using the following mathematical equation:

[tex]Distance = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}[/tex]

Where:

x and y represent the data points (coordinates) on a cartesian coordinate.

By substituting the given end points B (0, 2) and B' (5, 4) into the distance formula, we have the following;

[tex]Distance = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\\\\Distance \;BB'= \sqrt{(5-0)^2 + (4-2)^2}\\\\Distance \;BB'= \sqrt{(5)^2 + (2)^2}\\\\Distance \;BB'= \sqrt{25 + 4}\\\\Distance \;BB'= \sqrt{29}\;units[/tex]

Read more on distance here: brainly.com/question/12470464

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Answer:

Linear

Step-by-step explanation:

Took the test (USA Test prep)