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ratelena [41]
3 years ago
13

Two forces with magnitudes of 25 and 30 pounds act on an object at angles of 10° and 100°, respectively. Find the direction and

magnitude of the resultant force. Round to two decimal places in all intermediate steps and in your final answer.
I need an explanation please, I don't even know where to start.
Mathematics
1 answer:
Alina [70]3 years ago
6 0

Answer:

Here you go.

Step-by-step explanation:

x=25cos10+30cos100

y=25sin10+30sin100

v=√(x^2+y^2)

α=tan(y/x)  

Rounding to two decimal places in intermediate steps...

x≈19.41, y≈33.89

v≈39.05

α=60.20°

So (39.05, 60.20°)

Hope that helped, mark as brainliest asap not file exploiters

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A complex electronic system is built with a certain number of backup components in its subsystems. One subsystem has four identi
Tcecarenko [31]

Answer:

a. 0.1536

b. 0.9728

Step-by-step explanation:

The probability that a component fails, P(Y) = 0.2

The number of components in the system = 4

The number of components required for the subsystem to operate = 2

a. By binomial theorem, we have;

The probability that exactly 2 last longer than 1,000 hours, P(Y = 2) is given as follows;

P(Y = 2) = \dbinom{4}{2} × 0.2² × 0.8² = 0.1536

The probability that exactly 2 last longer than 1,000 hours, P(Y = 2) = 0.1536

b. The probability that the system last longer than 1,000 hours, P(O) = The probability that no component fails + The probability that only one component fails + The probability that two component fails leaving two working

Therefore, we have;

P(O) = P(Y = 0) + P(Y = 1) + P(Y = 2)

P(Y = 0) = \dbinom{4}{0} × 0.2⁰ × 0.8⁴ = 0.4096

P(Y = 1) = \dbinom{4}{1} × 0.2¹ × 0.8³ = 0.4096

P(Y = 2) = \dbinom{4}{2} × 0.2² × 0.8² = 0.1536

∴ P(O) = 0.4096 + 0.4096 + 0.1536 = 0.9728

The probability that the subsystem operates longer than 1,000 hours = 0.9728

4 0
2 years ago
Please help me with this D:
I am Lyosha [343]
Yes, because 120 - 36 = (120\times 36) - 4236 = 84

Hope that helped :D
8 0
3 years ago
at a party at the middle school over on third avenue next to the bank, $60\%$ of the students are seventh graders. these student
Vaselesa [24]

There are 600 students including the seventh and eighth graders at the party.

This problem uses the concept of percentages to define the conditions that are laid in front of us.

Let the original  number of students be S , and the number of seventh graders be   = 0.60S

We know that percent is used to convey the mathematical term of a fraction multiplied by 100.

Total students  after 20 eighth graders arrive = S + 20

And we have that

Number of seventh graders / total number of students  =  58%

.60S  / [ S + 20 ]  =   .58    we multiply both sides by  S +  20

0.60S  =0 .58 [ S + 20]

.60S  = .58S + 11.6   we subtract  0.58S from  both the sides

0.02S = 11.6  we divide both the sides  by .02

S  = 11/6 / 0.02 =    580

So the total number of students = 580 + 20  = 600 .

Hence there are 600 students at the party at that time.

To learn more about students visit:

brainly.com/question/17332524

#SPJ1

5 0
1 year ago
There are 9 computers and 72 students. what is the unit rate of students to computers?
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8 Is the unit rate. For 9/27
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