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Anna71 [15]
3 years ago
12

Solve the equation x=5/3πr^3 for r.​

Mathematics
1 answer:
JulsSmile [24]3 years ago
6 0

Answer:

go to symbolab and put that in, ur welcome

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once a week you babysit your neighbor’s toddler after school, usually going to a local playground. you notice that each swing on
Natalka [10]
You didn't include the formula.

Given that there is no data about the mass, I will suppose that the formula is that of the simple pendulum (which is only valid if the mass is negligible).

Any way my idea is to teach you how to use the formula and you can apply the procedure to the real formula that the problem incorporates.

Simple pendulum formula:

Period = 2π √(L/g)

Square both sides

Period^2 = (2π)^2 L/g

L = [Period / 2π)^2 * g

Period = 3.1 s
2π ≈ 6.28
g ≈ 10 m/s^2

L = [3.1s/6.28]^2 * 10m/s^2 =2.43 m

I  hope this helps you.
6 0
3 years ago
Sin2A+sin2B+sin2C÷4cosA/2.cosB/2.cosC/2​
Mrrafil [7]
The answer is c2 have a good day !!!! Bc you add all the numbers up and you will get it
3 0
2 years ago
Write an algebraic expression for: (1) 2 times a number squared increased by 6 ; -3
Over [174]

Answer:

Step-by-step explanation:

It appears i cant answer your question but what i can do is provide this paragraph as a trade for points. Appreciate your help. :)

5 0
3 years ago
Rewrite the equation by completing the square.<br> x^2+ 4x = 0
ExtremeBDS [4]

Answer:

Here you go

Step-by-step explanation:

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3 0
3 years ago
Read 2 more answers
A college student is taking two courses. The probability she passes the first course is 0.73. The probability she passes the sec
zhenek [66]

Answer:

b) No, it's not independent.

c) 0.02

d) 0.59

e) 0.57

f) 0.5616

Step-by-step explanation:

To answer this problem, a Venn diagram should be useful. The diagram with the information of Event 1 and Event 2 is shown below (I already added the information for the intersection but we're going to see how to get that information in the b) part of the problem)

Let's call A the event that she passes the first course, then P(A)=.73

Let's call B the event that she passes the second course, then P(B)=.66

Then P(A∪B) is the probability that she passes the first or the second course (at least one of them) is the given probability. P(A∪B)=.98

b) Is the event she passes one course independent of the event that she passes the other course?

Two events are independent when P(A∩B) = P(A) * P(B)

So far, we don't know P(A∩B), but we do know that for all events, the next formula is true:

P(A∪B) = P(A) + P(B) - P(A∩B)

We are going to solve for P (A∩B)

.98 = .73 + .66 - P(A∩B)

P(A∩B) =.73 + .66 - .98

P(A∩B) = .41

Now we will see if the formula for independent events is true

P(A∩B) = P(A) x P(B)

.41 = .73 x .66

.41 ≠.4818

Therefore, these two events are not independent.

c) The probability she does not pass either course, is 1 - the probability that she passes either one of the courses (P(A∪B) = .98)

1 - P(A∪B) = 1 - .98 = .02

d) The probability she doesn't pass both courses is 1 - the probability that she passes both of the courses P(A∩B)

1 - P(A∩B) = 1 -.41 = .59

e) The probability she passes exactly one course would be the probability that she passes either course minus the probability that she passes both courses.

P(A∪B) - P(A∩B) = .98 - .41 = .57

f) Given that she passes the first course, the probability she passes the second would be a conditional probability P(B|A)

P(B|A) = P(A∩B) / P(A)

P(B|A) = .41 / .73 = .5616

4 0
3 years ago
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