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Travka [436]
3 years ago
10

Given: F3 = 5 k + 515 g F4 = 945k - 713 g ; find F20

Mathematics
1 answer:
Alisiya [41]3 years ago
8 0

9514 1404 393

Answer:

  F20 = 15985k -20361g

Step-by-step explanation:

If we assume F3 and F4 are the third and fourth terms of an arithmetic sequence, the common difference is ...

  d = F4 -F3 = (945k -713g) -(5k +515g)

  d = 940k -1228g

Then the 20th term is ...

  F20 = F3 +(20 -3)d = (5k +515g) +17(940k -1228g)

  F20 = 15985k -20361g

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Answer: y=-2x-9

Step-by-step explanation:

If ANGL is a square, then NG and LG are adjacent sides.

Adjacent sides are perpendicular.  [Each angle is 90°]

The equation of line NG is Y=\dfrac12 X-6.

By comparing it to equation in slope intercept form y=mx+c ( where , m= slope , c=y-interecpt)

slope =\dfrac12

Let slope of LG be <em>n</em>, then

n\times \dfrac{1}{2}=-1 [Product of slopes of two perpendicular line =-1]

\Rightarrow n=-2

Equation of a line passes through (a,b) and have slope m is given by :-

(y-b)=m(x-a)

Equation of LG :

(y-1)=-2(x-(-5))\\\\\Rightarrow\ y-1=-2x-10\\\\\Rightarrow\ y=-2x-9 [In intercept form]

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3 years ago
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2 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

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40 percent 3/5 equals 60, 5/5 equals 100, 100-60 equals 40 percent

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