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4 years ago

I need help on this one

Mathematics

1 answer:

tia_tia [17]4 years ago

8 0

You need to come up with an array of answers for x and y that will be greater than 0.
x+4y>0. example if x is -1 and y is 1 will this be correct.
-1+(4*1)>0, -1+4>0. 3>0? Yes
if x is 1 and y is -1 then the equation becomes 1+(-4)>0 -3 is not>0 so it's not in the array
x could be -3 and y 1
You will plot a line to show all the answers.
i

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svetoff [14.1K]

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3 years ago

the graph shows the following equation. what are the solution to the system y = 4x + 21 and y =-(1/3)x+4+6

coldgirl [10]

Answer:

(-6,3) and (-4,-5) are the solutions to the system

Step-by-step explanation:

The solution to a system of equations are the point(s) where the graphes intersect. On the graph is a linear function and an exponential. Because of the curve in the exponential, the graph has two solutions at (-6,3) and (-4,-5).

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4 years ago

Suppose you flip a fair coin N times. Let random variable h be the number of heads that occur. Use the normal approximation to e

Mariana [72]

Answer:

If n = 1000000, then

$P(h E [495000, 505000]) = \int\limits^{50500}_{495000} {\frac{2}{\sqrt{2000000\pi}} e^{\frac{-(k-500000)^2}{500000} }} \, dk$

If n = 10400, then

$P(h E [495000, 505000]) = \int\limits^{50500}_{495000} {\frac{2}{\sqrt{2000000\pi}} e^{\frac{-(k-500000)^2}{500000} }} \, dk$

If N = 102, then

$P(h < 40 or h > 60) = 1-P(40$

Step-by-step explanation:

Since the coin is fair, then the probability that a filp is heads is 1/2. Given N tries, the amount of heads can be approximated with a Normal distribution with mean μ = N *1/2 = N/2 and standard deviation σ = √(N*1/2 * 1/2) = √N/ 2

The density function of that random variable is given by de following formula

$f_X(k) = \frac{1}{\sqrt{2\pi} * \sigma} e^{\frac{-(k-\mu)^2}{2\sigma^2} } = \frac{2}{\sqrt{2\pi N}} e^{\frac{-2(k-N/2)^2}{N} }$

If n = 1000000, then

$P(h E [495000, 505000]) = \int\limits^{50500}_{495000} {\frac{2}{\sqrt{2000000\pi}} e^{\frac{-(k-500000)^2}{500000} }} \, dk$

If n = 10400, then

$P(h E [495000, 505000]) = \int\limits^{50500}_{495000} {\frac{2}{\sqrt{2000000\pi}} e^{\frac{-(k-500000)^2}{500000} }} \, dk$

If N = 102, then

$P(h < 40 or h > 60) = 1-P(40$

4 0

3 years ago

What is the result when 16x + 9 is subtracted from 4x + 6

romanna [79]

Answer:

(4x+9)-(16x+9) = 4x+9-16x-9

= -12x

6 0

3 years ago

Car X weighs 136 pounds more than car Z. Car Y weighs 117 pounds more than car Z. The total weight of all three cars is 9439 pou

Aleksandr [31]

Let x, y and z denote the weighs of car X, car Y and car Z, respectively.

We know that car X weighs 136 more than car Z, this can be express by the equation:

$x=z+136$

We also know that Y weighs 117 pounds more than car Z, this can be express as:

$y=z+117$

Finally, we know that the total weight of all the cars is 9439, then we have:

$x+y+z=9439$

Hence, we have the system of the equations:

$\begin{gathered} x=z+136 \\ y=z+117 \\ z+y+z=9439 \end{gathered}$

To solve the system we can plug the values of x and y, given in the first two equations, in the last equation; then we have:

$\begin{gathered} z+136+z+117+z=9439 \\ 3z=9439-136-117 \\ 3z=9186 \\ z=\frac{9186}{3} \\ z=3062 \end{gathered}$

Now that we have the value of z we plug it in the first two equations to find x and y:

$\begin{gathered} x=3062+136=3198 \\ y=3062+117=3179 \end{gathered}$

Therefore, car X weighs 3198 pound, car Y weighs 3179 pounds and car Z weighs 3062 pounds.

4 0

1 year ago