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Marat540 [252]
3 years ago
11

Solve for n: c= 6n−5g /11t

Mathematics
1 answer:
qwelly [4]3 years ago
3 0

Answer: \frac{11tc+5g}{66t}=n

Step-by-step explanation:

You have the equation c=6n-\frac{5g}{11t}.

Then, to solve for the variable n from the equation you need:

Make the subtraction of the right side of the equation:

(As the denominators are 1 and 11t, the least common denominator is 11t)

c=\frac{(6n)(11t)-5g}{11t}\\\\c=\frac{66nt-5g}{11t}

Multiply  11t to both sides:

(11t)c=(\frac{66nt-5g}{11t})(11t)\\\\11tc=66nt-5g

Add 5g to both sides:

11tc+5g=66nt-5g+5g\\\\11tc+5g=66nt

And finally divide both sides by 66t:

\frac{11tc+5g}{66t}=\frac{66nt}{66t}\\\\\frac{11tc+5g}{66t}=n

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KengaRu [80]

Answer:

B

Step-by-step explanation:

8 0
3 years ago
The probability that a randomly selected 2 2​-year-old male garter snake garter snake will live to be 3 3 years old is 0.98861 0
Mnenie [13.5K]

Answer:

a. Probability = 0.97735

b. Probability = 0.92294

c. P(At\ Least\ One) = 1

No, it is not unusual if at least 1 lives up to 3.

Step-by-step explanation:

Given

Represent the probability that a 2 year old snake will live to 3 with P(Live);

P(Live) = 0.98861

Solving (a): Probability that two selected will live to 3 years.

Both snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

Probability = P(Live)\ and\ P(Live)

Probability = 0.98861 * 0.98861

Probability = 0.9773497321

Probability = 0.97735 <em>--- Approximated</em>

Solving (b): Probability that seven selected will live to 3 years.

All 7 snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

Probability = P(Live)^n

Where n = 7

Probability = 0.98861^7

Probability = 0.92294324145

Probability = 0.92294 <em>--- Approximated</em>

Solving (c): Probability that at least one of seven selected will not live to 3 years.

In probabilities, the following relationship exist:

P(At\ Least\ One) = 1 - P(None).

So, first we need to calculate the probability that none of the 7 lived up to 3.

If the probability that one lived up to 3 years is 0.98861, then the probability than one do not live up to 3 years is 1 - 0.98861

This gives:

P(Not\ Live) = 0.01139

The probability that none of the 7 lives up to 3 is:

P(None) = P(Not\ Live)^7

P(None) = 0.01139^7

Substitute this value for P(None) in

P(At\ Least\ One) = 1 - P(None).

P(At\ Least\ One) = 1 - 0.01139^7

P(At\ Least\ One) = 0.99999999999997513055642436060443621

P(At\ Least\ One) = 1 ---- Approximated

No, it is not unusual if at least 1 lives up to 3.

This is so because the above results, which is 1 shows that it is very likely for at least one of the seven to live up to 3 years

7 0
3 years ago
What are all the answers to this screenshot
sp2606 [1]

Answer:

b. 0.1 difference

c. 1.04 times

d. 6th, 8th, 7th

7 0
3 years ago
What is the answer to 5x+4=9
Kisachek [45]

Answer:

x=1

Step-by-step explanation:

5x-4=9

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x=1

8 0
3 years ago
A white label wrapped around a rectangular box with square bases covers 80 percent of the lateral surface area of the box. Find
erastova [34]
80% = 6 in
100%=?


100/x 80/6 (proportional relationship)

then cross multiply

600 = 80x
600/80= 7.5

x=7.5
4 0
3 years ago
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