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weqwewe [10]
3 years ago
7

Please help me out!

Mathematics
1 answer:
Lesechka [4]3 years ago
8 0

Answer:

0

Step-by-step explanation:

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Divide the equation and provide a step through process
Allushta [10]

Answer:

300x^12/15x^10=300/15*x^12/x^10=20*x^2=20x^2

a^4/a^4=1

g^6/g^10=g^-4=1/g^4

so the answer is (20x^2)/g^4

Step-by-step explanation:

5 0
2 years ago
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Help!!!!!!!!!!?.........
raketka [301]

Answer:

I think E.

Step-by-step explanation:

5 0
3 years ago
A series of quarterly payments of P1,300 each at the first payment is due at 4 years, and the last payment at the end of 12 year
gtnhenbr [62]

Answer:

The Value of Payments is  P18,557.15

Step-by-step explanation:

The quarterly payment is an annuity payment.

Use the following formula to calculate the present value of the payments.

PV of Annuity = Annuity Payment x ( 1 - ( 1 + interest rate )^-Numbers of periods ) / Interest rates

Where

Annuity Payment = Quarterly payment = P1,300

Interest rate = 5.12% x 3/12 = 1.375%

Numbers of periods = 4 years x 12/3 = 16 quarters

PV of Annuity = Value of Payments  = ?

Placing values in the formula

Value of Payments = P1,300 x ( 1 - ( 1 + 1.375% )^-16 ) / 1.375%

Value of Payments =  P18,557.15

4 0
2 years ago
6. Assume that a component passes a test is 0.85 and that components perform independently. What is the probability that the thi
Tanzania [10]

Answer:

3.90% probability that the third failure will occur on the tenth component tested

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it fails, or it does not fail. Components perform independently. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Assume that a component passes a test is 0.85

So they fail with probability of p = 1 - 0.85 = 0.15

What is the probability that the third failure will occur on the tenth component tested

First 9 components: Two failures, that is, P(X = 2) when n = 9.

10th component: Failure with probability 0.15.

So

P = 0.15P(X = 2)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{9,2}.(0.15)^{2}.(0.85)^{7} = 0.2597

So

P = 0.15P(X = 2) = 0.15*0.2597 = 0.0390

3.90% probability that the third failure will occur on the tenth component tested

5 0
3 years ago
Find the valie when x=2 and y=3 2x^0 y^-2 a. 1/9 b.2/9 c.-18​
sp2606 [1]

Answer:

b

x^0=1

y^-2 =1/y^2

Step-by-step explanation:

2x^0 y^-2

2×1 ×1/3^2

2×1/9

2/9

4 0
3 years ago
Read 2 more answers
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