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If the table represents a linear function, what's the missing value of y?

Svetach [21]

C) 13

The y value adds one by every x; therefore 4 would be 13.

3 0

2 years ago

Write an equation in point slope and slope intercept form of a line that passes through the given point and

nekit [7.7K]

Answer:

Point slope is ( Y+4) = 1/2(x+3)

Slope intercept is Y = 1/2(x) -5/2

Step-by-step explanation:

For the point slope form.

Given the point as (-3,-4)

And the gradient m = 1/2

Point slope form is

(Y - y1) = m(x-x1)

So

X1 = -3

Y1 = -4

(Y - y1) = m(x-x1)

(Y - (-4)) = 1/2(x -(-3))

( Y+4) = 1/2(x+3)

For the slopes intercept form

Y = mx + c

We can continue from where the point slope form stopped.

( Y+4) = 1/2(x+3)

2(y+4)= x+3

2y + 8 = x+3

2y = x+3-8

2y = x-5

Y = x/2 - 5/2

Y = 1/2(x) -5/2

Where -5/2 = c

1/2 = m

7 0

2 years ago

Suppose the average price of gasoline for a city in the United States follows a continuous uniform distribution with a lower bou

Novay_Z [31]

Answer:

$P(X$

And we can use the cumulative distribution function given by:

$F(x) = \frac{x-a}{b-a} , a \leq X \leq b$

And for this case we can write the probability like this:

$P(X$

And then the final answer for this case would be $\frac{2}{3}=0.667$

Step-by-step explanation:

For this case we define our random variable X "price of gasoline for a city in the USA" and we know the distribution is given by:

$X \sim Unif (a=3.5, b=3.8)$

And for this case the density function is given by:

$f(x) = \frac{x}{b-a}= \frac{x}{3.8-3.5}=, 3.5 \leq X \leq 3.8$

And we want to calculate the following probability:

$P(X$

And we can use the cumulative distribution function given by:

$F(x) = \frac{x-a}{b-a} , a \leq X \leq b$

And for this case we can write the probability like this:

$P(X$

And then the final answer for this case would be $\frac{2}{3}=0.667$

5 0

2 years ago

A landscaper wants to create a 12-foot-long diagonal path through a rectangular garden. The width of the garden is x feet and th

dolphi86 [110]

Answer:

Width of the rectangle = 6.7 ft

length of the rectangle = 10.7 ft

Step-by-step explanation:

ABCD is the rectangle.

AB = length of the rectangle = 4 + x ft

BC = width of the rectangle = x ft

AC = Diagonal of the rectangular field = 12 ft

Since ΔABC is the Right angle triangle. So

$AC^{2} = AB^{2} + BC^{2}$

$12^{2} = (4 + x)^{2} + x^{2}$

$144 = x^{2} + 16 + x^{2} + 8 x$

$144 = 2x^{2} + 8x + 16$

$x^{2} + 4x -72 = 0$

By solving above equation we get

x = 6.7 ft

Thus is the width of the rectangle.

And length of the rectangle = 4 + x

⇒ 4 + 6.7

⇒ 10.7 ft

5 0

2 years ago

Read 2 more answers

Solve y′′=sin(x) if y(0)=0 and y′(0)=5.<br><br> y(x)=?

Ksenya-84 [330]

Answer:

$y(x)=6x-sin(x)$

Step-by-step explanation:

Rewrite the differential equation as:

$\frac{d^{2} y }{dx^{2} } =sin(x)$

Integrate both sides with respect to x:

$\int\ \frac{d^{2} y }{dx^{2} } dx = \int\ sin(x) dx$

$\frac{dy}{dx} =-cos(x)+C_1$

Integrate one more time both sides with respect to x:

$\int\ \frac{dy}{dx} = \int\ -cos(x)+C_1 dx$

$y(x)=-sin(x)+C_1x+C_2$

Now that we find the solution, let's find its derivate:

$y'(x)=C_1-cos(x)$

Evaluating the initial conditions:

$y(0)=C_1(0)+C_2-sin(0)=0\\C_2=0$

$y'(0)=C_1-cos(0)=5\\C_1=5+1=6$

Replacing the value of the constants that we found in the differential equation solution:

$y(x)=6x-sin(x)$

6 0

2 years ago