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zubka84 [21]
3 years ago
8

What is 180(n-2)=s when solving for N​

Mathematics
1 answer:
Sergio039 [100]3 years ago
5 0

Answer:

n = s/180 +2

Step-by-step explanation:

180(n-2)=s

Distribute

180n - 360 =s

Add 360 to each side

180n -360+360 = s+360

180n = s+360

Divide each side by 180

180n/180 = s/180 + 360/180

n = s/180 +2

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37800 meters

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3600 seconds in an hour x 3

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multiply that number by 7, days a week

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3 years ago
What is 817x45 in standerd algorithm pls help
sineoko [7]

Answer:

Step-by-step explanation:

  817

x 45

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36,765

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3 years ago
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The population of a certain animal species you are studying decreases at a rate of 3.5% per year. You have counted 80 of the ani
kodGreya [7K]

Answer:

  65

Step-by-step explanation:

Next year, there will be 3.5% fewer, so 100% - 3.5% = 0.965 times the number present this year. This same multiplication is repeated each year.

After 6 years, the number present will be 0.965^6 times the number this year, or ...

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There will be about 65 animals after 6 years.

3 0
3 years ago
Sandy plays tennis once every 6 six days. Jim plays once every 8 eight days. If they both played tennis today, in how many days
tiny-mole [99]

Answer: 24 days

Step-by-step explanation:

Sandy plays tennis once every 6 six days while Jim plays tennis once every 8 eight days.

To get the number of days they will play on the same day again, we find the lowest common multiple for the numbers.

The lowest common multiple is the smallest positive integer which is evenly divisible by both 6 and 8.

6 = 6, 12, 24, 30

8 = 8, 16, 24, 32

The lowest common multiple is 24.

In 24 days, they will play on the same day again

8 0
4 years ago
You fire 3 guns at a target. the first has a 20% probability of hitting, the second a 15% probability of hitting and the 3rd an
DedPeter [7]
<span>Sample space for hits: {0, 1, 2, 3} Expected hits: 0.45 Variance: 0.3775 We have 3 guns, each of which can either hit or miss. The sample space for the number of hits is {0, 1, 2, 3} The expected number of hits is 0.20 + 0.15 + 0.10 = 0.45 hits For the variance, let's calculate the probability of 0, 1, 2, or 3 hits. 0 hits: 0.80 * 0.85 * 0.90 = 0.612 1 hit: 0.8 * 0.85 * 0.1 + 0.8 * 0.15 * 0.9 + 0.2 * 0.85 * 0.9 = 0.068 + 0.108 + 0.153 = 0.329 2 hits: 0.8 * 0.15 * 0.1 + 0.2 * 0.85 * 0.1 + 0.2 * 0.15 * 0.9 = 0.012 + 0.017 + 0.027 = 0.056 3 hits: 0.2 * 0.15 * 0.1 = 0.003 Since we have 4 discrete outcomes, the variance is simply the sum of the probability of the event multiplied by the square of the difference between the value of the event minus the mean. So 0.612 * (0 - 0.45)^2 + 0.329 * (1 - 0.45)^2 + 0.056 * (2 - 0.45)^2 + 0.003 * (3 - 0.45)^2 = 0.612 * (-0.45)^2 + 0.329 * (0.55)^2 + 0.056 * (1.55)^2 + 0.003 * (2.55)^2 = 0.612 * 0.2025 + 0.329 * 0.3025 + 0.056 * 2.4025 + 0.003 * 6.5025 = 0.12393 + 0.0995225 + 0.13454 + 0.0195075 = 0.3775 So the variance is 0.3775 Note: I could have eliminated the step consolidating the probabilities of getting 0, 1, 2, or 3 hits and instead went directly into the calculation of the variance using the probability of each of the 8 possible events. But in doing so that would have made the calculations twice as large and still would have gotten the same effect. Effectively for the cases of 1 and 2 hits, I calculated (p1 + p2 + p3)*(1 - 0.45)^2 instead of p1*(1 - 0.45)^2 + p2*(1 - 0.45)^2 + p3*(1 - 0.45)^2 which the distributive property allows us and will give the same result.</span>
6 0
3 years ago
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