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3 years ago

Given that f(x) = 2x – 1, find f(6). Your answer: 6 7 11 13

Mathematics

2 answers:

ira [324]3 years ago

8 0

Input interpretation:

2 x - 1 where x = 6

Result:

Kruka [31]3 years ago

4 0

Answer:

Step-by-step explanation:

2 times 6 is 12 minus 1 is 11

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Find the value of each expression.

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I'm not sure

Step-by-step explanation:

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2 years ago

4)Without a calculator, name the zeros and sketch the graph.<br> 56.) = -2x(x-4)^2(x+3)^2

Ilia_Sergeevich [38]

Answer:

0, 4, -3

Step-by-step explanation:

replace y with 0 and then fro the sketch I recommend you tomuse desmos good luck buddie

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3 years ago

If the given figure is rotated 180° around the origin, what are the new coordinates of point Z? Z= (6,-9)

Nitella [24]

Answer:

Step-by-step explanation:

Rule for 180 rotation: (x,y) becomes (-x,-y)

So (6,-9) becomes (-6,9)

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3 years ago

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Blizzard [7]

Answer:

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Step-by-step explanation:

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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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3 years ago