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poizon [28]
2 years ago
9

36x-6 distributive property

Mathematics
1 answer:
Ahat [919]2 years ago
3 0

Answer:

36x-6=0

36x-6+6=0+6

36x=0+6

36x=6

x=6÷36

x=6

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1.01<0.99 true,4.5=4.50 true, 3.5<3.39 true?, lastly is 1.51>1.15 true?
kirza4 [7]
No, yes, no, yes

What part of comparing number values do you not understand? Most folks figure out enough about place value to be able to answer this by 3rd grade. If there's something about this question that really stumps you, please advise in the comments.
5 0
3 years ago
The owner of an automobile insures it against damage by purchasing an insurance policy with a deductible of 250. In the event th
choli [55]

Answer:

Step-by-step explanation:

From the given information:

The uniform distribution can be represented by:

f_x(x) = \dfrac{1}{1500} ; o \le x \le   \  1500

The function of the insurance is:

I(x) = \left \{ {{0, \ \ \ x \le 250} \atop {x -20 , \ \  \ \ \ 250 \le x \le 1500}} \right.

Hence, the variance of the insurance can also be an account forum.

Var [I_{(x}) = E [I^2(x)] - [E(I(x)]^2

here;

E[I(x)] = \int f_x(x) I (x) \ sx

E[I(x)] = \dfrac{1}{1500} \int ^{1500}_{250{ (x- 250) \ dx

= \dfrac{1}{1500 } \dfrac{(x - 250)^2}{2} \Big |^{1500}_{250}

\dfrac{5}{12} \times 1250

Similarly;

E[I^2(x)] = \int f_x(x) I^2 (x) \ sx

E[I(x)] = \dfrac{1}{1500} \int ^{1500}_{250{ (x- 250)^2 \ dx

= \dfrac{1}{1500 } \dfrac{(x - 250)^3}{3} \Big |^{1500}_{250}

\dfrac{5}{18} \times 1250^2

∴

Var {I(x)} = 1250^2 \Big [ \dfrac{5}{18} - \dfrac{25}{144}]

Finally, the standard deviation  of the insurance payment is:

= \sqrt{Var(I(x))}

= 1250 \sqrt{\dfrac{5}{48}}

≅ 404

4 0
3 years ago
What is the equation of the line with a slope of –3 and a y-intercept of 4?
arsen [322]

Answer:

y=-3x+4

y=mx+b

m is the slope

b is the yint

5 0
3 years ago
Read 2 more answers
Find the slope between:<br> (4, -1)<br> (-2,4)
Natali5045456 [20]

Answer:

-5

----

6

Step-by-step explanation:

y2-y1

-------

x2-x1

-1-4 -5

------ = ---

4--2 6

~Anna

6 0
3 years ago
Question 5 of 30<br> What is the period of the graph of y<br> 5cos(4x – 1) + 3?
Anastaziya [24]

Answer:

Period is \frac{\pi}{2}

Step-by-step explanation:

For the function y=acos(bx-c)+d, the period is \frac{2\pi}{|b|}.

Therefore, the period of the function is \frac{2\pi}{|b|}=\frac{2\pi}{|4|}=\frac{2\pi}{4}=\frac{\pi}{2}.

What this means is that the function's wave cycle will repeat every \frac{\pi}{2} units.

8 0
3 years ago
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