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agasfer [191]
3 years ago
12

7 3 If you were to graph a line with a slope of 3/2 that intercepts the y-axis at -2 on the coordinate plane, one point on that

line would have a y coordinate of 4 with an x coordinate of what number?​
Mathematics
1 answer:
pychu [463]3 years ago
3 0

Answer:

4

Step-by-step explanation:

Make a graph.

Start at negative 2 because that's the y intercept. Remember that a minus sign in front of the y intercept just means it's negative.

The slope is basically the rise over the run. How much you go UP divided by how much you run. Since the slope is 3/2, we go up 3 and right 2.

Keep making points and you get 4

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A small tanker truck can hold 210 gallons of heating oil. How many pints of heating oil can the truck hold
shtirl [24]

Answer: 1,680 pints

Step-by-step explanation: 210 gallons= 1,680 pints

8 0
4 years ago
2 A cafe sells cakes and scones.
Alecsey [184]
30 cakes on tuesday
8 0
3 years ago
In a family, the father took 1/5 of the cake and he had 4 times as much as others had, then the family members are________. A. 1
kramer
There is one cake
It is divided among x family members, so each gets:
1/x
The father took 1/5 which is 4 times greater than each one's piece. So:
1/5 = 4(1/x)
1/5 = 4/x
x = 20
There are 20 family members so the answer is none of these
5 0
4 years ago
Suppose X and Y are random variables with joint density function. f(x, y) = 0.1e−(0.5x + 0.2y) if x ≥ 0, y ≥ 0 0 otherwise (a) I
Hatshy [7]

a. f_{X,Y} is a joint density function if its integral over the given support is 1:

\displaystyle\int_{-\infty}^\infty\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\frac1{10}\int_0^\infty\int_0^\infty e^{-x/2-y/5}\,\mathrm dx\,\mathrm dy

=\displaystyle\frac1{10}\left(\int_0^\infty e^{-x/2}\,\mathrm dx\right)\left(\int_0^\infty e^{-y/5}\,\mathrm dy\right)=\frac1{10}\cdot2\cdot5=1

so the answer is yes.

b. We should first find the density of the marginal distribution, f_Y(y):

f_Y(y)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx=\frac1{10}\int_0^\infty e^{-x/2-y/5}\,\mathrm dy

f_Y(y)=\begin{cases}\dfrac15e^{-y/5}&\text{for }y\ge0\\\\0&\text{otherwise}\end{cases}

Then

P(Y\ge8)=\displaystyle\int_8^\infty f_Y(y)\,\mathrm dy=e^{-8/5}

or about 0.2019.

For the other probability, we can use the joint PDF directly:

P(X\le5,Y\le8)=\displaystyle\int_0^5\int_0^8f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=1+e^{-41/10}-e^{-5/2}-e^{-8/5}

which is about 0.7326.

c. We already know the PDF for Y, so we just integrate:

E[Y]=\displaystyle\int_{-\infty}^\infty y\,f_Y(y)\,\mathrm dy=\frac15\int_0^\infty ye^{-y/5}\,\mathrm dy=\boxed5

5 0
3 years ago
One month ruby wokred 6 more hours than isaac and svetlana worked 4 times as many as ruby they together worked 126 hours how man
jok3333 [9.3K]
R + R - 6 + 4R = 126
6R - 6 = 126
6R = 126 + 6
6R = 132
R = 132/6
R = 22 (hours Ruby worked)
I = R - 6
I = 22 - 6
I = 16 (hours Isaac worked)
S = 4R
S = 4(22)
S = 88 (hours Svetlana worked)

Finally, let's check our three answers to see if their sum is 126 (the given total of hours worked):
22 + 16 + 88 = 38 + 88 = 126


8 0
3 years ago
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