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In the given figure AOB is a straight line find the value of x. Please tell the correct answer..

Scilla [17]

Answer:

2x+5x+3x=180°

10x=180°

x=180/10

x=18° ans..

5 0

3 years ago

Read 2 more answers

Show that y=cos(t)y=cos(t) is a solution to (dydt)2=1−y2(dydt)2=1−y2. Enter your answers below in terms of the independent varia

GrogVix [38]

Answer:

$(\frac{dy}{dt})^2=sin^2t$

Step-by-step explanation:

$y=cost$

DE : $(\frac{dy}{dt})^2=1-y^2$

If y is a solution of given DE then it satisfied the DE.

Differentiate w.r.t t

$\frac{dy}{dt}=-sint$

Using the formula

$\frac{d(cosx)}{dx}=-sinx$

LHS: $(\frac{dy}{dt})^2=(-sint)^2=sin^2t$

RHS

$1-y^2=1-cos^2t=sin^2t$

By using the formula

$sin^2t=1-cos^2t$

LHS=RHs

Hence, y is a solution of given DE

$(\frac{dy}{dt})^2=sin^2t$

7 0

3 years ago

A woman in a highland village in the Andes knits sweaters and sells them for export. She also takes care of her family and helps

Mnenie [13.5K]

Answer:

Expected number of sweaters per month can be given as follows:

E(X) = Σ x P(X = x)

Now,

E(X) = [2 * 0.1 + 3* 0.1+ 4* 0.2 + 5* 0.3 + 6* 0.2 + 7 * 0.1]

E(X) = 4.7.

Var(X) = E(X^2) – [E(X)]^2

We have E(X) = 4.7. Thus, [E(X)]2= 4.7*4.7 = 22.09.

Now E(X^2) = [2*2 * 0.1 + 3*3* 0.1+ 4*4* 0.2 + 5*5* 0.3 + 6*6* 0.2 + 7*7*0.1]

E(X^2) = 24.1

Thus by formula, Var(X) = E(X2) – [E(X)]2

Var(X) = 24.1-22.09

Var(X) = 2.01

Given that exporter pays the $12 for each sweater. The woman pays $2 per sweater. The cost of shipment is $3 irrespective of the number of sweaters. Now, let m is the number of sweaters she made. Thus, the total cost she would have to pay would be

Total cost by woman = 2m+3

The total cost paid by the exporter would be = 12m.

Now the profit of woman would be given by,

= The total cost exporter pay – cost paid by the woman

= 12m – (2m +3)

= 12m – 2m -3

= 10m – 3.

Now expected profit made by the woman is given in the following table below:

E(Profit) = Σ profit* P(X = x)

In a similar way, as we have done in part (a).

E(Profit) = [17 * 0.1 + 27* 0.1+ 37* 0.2 + 47* 0.3 + 57* 0.2 + 67 * 0.1]

E(Profit) = 44.

Now, we calculate the variance:

Var(profit) = E(profit^2) – [E(profit)]^2

Var(profit) =

E(profit^2) = [17*17 * 0.1 + 27*27* 0.1+ 37*37* 0.2 + 47*47* 0.3 + 57*57* 0.2 + 67*67*0.1]

E(profit^2) = 2137.

[E(profit)]^2 = 44*44 = 1936.

Thus, the variance can be given as =

Var(profit)= 2137 – 1936

Var(profit) = 201.

6 0

2 years ago

Please help me! (this is from khan academy 6th grade math!)

Svetach [21]

Answer:

1st → 1st

2nd → 3rd

3rd → 2nd

Step-by-step explanation:

just divide the first set of student wishes to the second set of students in the same question. the 1st for example

x-ray to all students is 6/ 36= 1/6 = 1:6

for all students, you add all the students in the table

8 0

1 year ago

A certain region currently has wind farms capable of generating a total of 2500 megawatts (2.5 gigawatts) of power. Complete p

marishachu [46]

Answer:

The correct answer is A. 7,665'000,000 kilowatt-hours per year and B. 766,500 households.

Step-by-step explanation:

1. Let's review the information provided to us for solving the questions:

Power capacity of the wind farms = 2,500 Megawatts or 2.5 Gigawatts

2. Let's resolve the questions a and b:

Part A

Assuming wind farms typically generate 35% of their capacity, how much energy, in kilowatt-hours, can the region's wind farms generate in one year?

2,500 * 0.35 = 875 Megawatts

875 Megawatts = 875 * 1,000 Kilowatts = 875,000 Kilowatts

Now we calculate the amount of Kilowatts per hour, per day and per year:

875.000 Kw generated by the farms means that are capable of produce 875,000 kw per hour of energy

875,000 * 24 = 21'000,000 kilowatt-hours per day

21'000,000 * 365 = 7,665'000,000 kilowatt-hours per year

Part B

Given that the average household in the region uses about 10,000 kilowatt-hours of energy each year, how many households can be powered by these wind farms?

For calculating the amount of households we divide the total amount of energy the wind farms can generate (7,665'000,000 kilowatt-hours) and we divide it by the average household consumption (10,000 kilowatt-hours)

Amount of households = 7,665'000,000/10,000 = 766,500

4 0

2 years ago

I need help with this​

I need help with this