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Studentka2010 [4]

3 years ago

13

Race car driver Mario won 3 races and earn 75 points for each win. During the fourth race, he had dedications of 55 points, 104

points, and 85 points, but earned 3 points for finishing. Write and find the value of an expression for how many points he has now

1 answer:

svet-max [94.6K]3 years ago

5 0

Answer:

he has 472 points now.

Step-by-step explanation:

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I need help with question #2. ASAP. Anything is appreciated:)

DedPeter [7]

Y = 4/5x + 30

4/5 because the slope is 4 up, 5 right. 30 is the y-int

4 0

3 years ago

Need some help, please. What do I plot? I attempted to figure the questions out many times, but I've come to no clear conclusion

olganol [36]

Answer:

vertex is (4,-4) and another point is (6,0) or you could use (2,0) or many other options :)

Step-by-step explanation:

The cool thing about this question your quadratic is in factored form so your x-intercepts are easy to figure out, they are 2 and 6.

So you can plot (6,0) and (2,0).

The vertex will lie half between x=2 and x=6... so it lays at (6+2)/2=4

We just have to find the y-coordinate for when x=4.

Plug in 4 gives you (4-2)(4-6)=(2)(-2)=-4.

So the vertex is at (4,-4).

5 0

3 years ago

Suppose that the length of a side of a cube X is uniformly distributed in the interval 9

Nastasia [14]

Answer:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

Step-by-step explanation:

Given

$9 < x < 10$ --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

$v = x^3$

For a uniform distribution, we have:

$x \to U(a,b)$

and

$f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.$

$9 < x < 10$ implies that:

$(a,b) = (9,10)$

So, we have:

$f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Solve

$f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Recall that:

$v = x^3$

Make x the subject

$x = v^\frac{1}{3}$

So, the cumulative density is:

$F(x) = P(x < v^\frac{1}{3})$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$ becomes

$f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.$

The CDF is:

$F(x) = \int\limits^{v^\frac{1}{3}}_9 1\ dx$

Integrate

$F(x) = [v]\limits^{v^\frac{1}{3}}_9$

Expand

$F(x) = v^\frac{1}{3} - 9$

The density function of the volume F(v) is:

$F(v) = F'(x)$

Differentiate F(x) to give:

$F(x) = v^\frac{1}{3} - 9$

$F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}$

$F'(x) = \frac{1}{3}v^{-\frac{2}{3}}$

$F(v) = \frac{1}{3}v^{-\frac{2}{3}}$

So:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

8 0

2 years ago

Help will give Brainly points

Elodia [21]

Answer:

$\frac{24}{25}$

Step-by-step explanation:

$\cos(\theta)=\frac{\textrm{adjacent}}{\textrm{hypotenuse}}$

4 0

2 years ago

Is 3/3 and 6/6 equivalent

ella [17]

Yes. Because both equals 1 whole.

7 0

2 years ago

Read 2 more answers