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Nastasia [14]
2 years ago
10

Anyone out there to help me? Please.

Mathematics
1 answer:
Marta_Voda [28]2 years ago
5 0

Answer:

x=72°

Step-by-step explanation:

A=360-102-95-55=108

x=180-108=72

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You are thinking of using a t procedure to construct a 95% confidence interval for the mean of a population. You suspect the dis
bonufazy [111]

this makes no sense sos im only 8 years old.

5 0
2 years ago
Which of the following square roots would not be between 4 and 5?
expeople1 [14]

Answer:

√10.

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7 0
2 years ago
Read 2 more answers
The table below shows some inputs and outputs of the invertible function f with domain all real numbers.
Ratling [72]

Answer:

{f}^{ - 1} (f(93)) = 93

f(f(7)) =  - 15

Step-by-step explanation:

A function and its inverse has the following properties

{f}^{ - 1} (f(x)) = x

This implies that;

{f}^{ - 1} (f(93)) = 93

From the table,

f(7) =  - 6

This means that:

f(f(7)) = f( - 6)

f(f(7)) =  - 15

Note that: f(7) means the value of f(x) when x=7.

From the table f(x)=-6 when x=7.

That is why f(7)=-6.

5 0
3 years ago
Please help me I don't know how to do it
poizon [28]
Findd total collected
add everybody
fri+sat+sun+mon=
21+5 and 5/6+2.75+4 and 5/8
add what you can
21+5+2.75+4+5/6+5/8=
28.75+5/6+5/8
add the fractions by making denom same
5/6+5/8=20/24+15/24=35/24=1 and 9/24=1 and 3/8
28.75+1 and 3/8=29.75+3/8
that is how many barels total
times 205 since that is how many litesr in each bottle
205 times (29.75+3/8)=
205*29.75+205*(3/8)=
6098.75+76.875=
6175.625

29/30 is lost

1//30 remains
9175.625 times 1/30 (aka divided by 30)=205.854166666666666 litesr
205 bottles (and 0.85 partially full bottle)
3 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
2 years ago
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