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3 years ago

The 12th term of GP whose 1 first term is 1/8 and second term is 1/2is

Mathematics

1 answer:

Inessa05 [86]3 years ago

3 0

Answer:

jjanation:jdgjdjgdjgjkdkidjgjghdjjghhkd

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Solve the equation X=

4vir4ik [10]

Step-by-step explanation:

I think the answer is x=70

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2 years ago

How do you solve secx-2=0

siniylev [52]

It will be secx = 2
or, cosx = 1/2
or x = Π/3 , 5Π/3

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3 years ago

Find the area of the triangle below.

Rom4ik [11]

Step-by-step explanation:

the area of a triangle is

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in our case we have a baseline with its associated height right there. and so the area is

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2 years ago

2.) Bob has two weekend jobs. Last weekend he made a total of $77 after

Butoxors [25]

Answer:

Step-by-step explanation:

x= money earned per hour working as a cashier.

y=money earned per hour working delivering newspapers.

We propose the following system of equations:

5x + 4y =77

6x + 3y=78

We solve the system by reduction method:

-6*(5x+4y=77) ⇒ -30x-24y=-462

5*(6x+3y=78)⇒ 30x+15y=390

---------------------------

-9y=-72 ⇒ y=-72 /-9=8

Now, we get the value of "x" replacing the value of "y" by "8" in any part of the equation above.

5x + 4(8)=77

5x+32=77

5x=77-32

5x=45

x=45/5=9

therefore;

money earned per hour working as a cashier= $9/ hour

money earned per hour working delivering newspapers=$8/hour

7 0

3 years ago

Read 2 more answers

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