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kotykmax [81]
3 years ago
5

Part 4. Find the missing sides for the 30 60 90 triangles. Please show work for each​

Mathematics
2 answers:
inna [77]3 years ago
7 0

10. tan(30) = l / 6√3

l = tan(30) x 6√3

l = 6

cos(30) = 6√3 / y

y =  6√3/cos(30)

y = 12

11. tan(30) = c/30

c = 30 x tan(30)

c = 10√3

cos(30) = 27/w

w = 27/cos(30)

w = 18√3

12.

sin(60) = 36/y

y = 36/sin(60)

y = 24√3

tan(60) = 36/x

x = 36/tan(60)

x = 12√3

tatuchka [14]3 years ago
3 0

Answer:

  10. (l, y) = (6, 12)

  11. (c, w) = (9√3, 18√3)

  12. (x, y) = (12√3, 24√3)

Step-by-step explanation:

Side ratios, shortest to longest, in a 30°-60°-90° triangle are ...

  1 : √3 : 2

__

10. l : 6√3 : y = 1 : √3 : 2

multiply by 6

  = 6 : 6√3 : 12

  l = 6, y = 12

__

11. c : 27 : w = 1 : √3 : 2

multiply by 27/√3 = 9√3

  = 9√3 : 27 : 18√3

  c = 9√3, w = 18√3

__

12. x : 36 : y = 1 : √3 : 2

multiply by 36/√3 = 12√3

  = 12√3 : 36 : 24√3

  x = 12√3, y = 24√3

__

This is easier when the triangle is drawn to clearly differentiate the lengths of the sides. The long leg is opposite the 60° angle; the short leg is opposite the 30° angle.

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(a) The probability that a randomly selected time interval between eruptions is longer than 82 ​minutes is 0.3336.

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(c) The probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 ​minutes is 0.0055.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) The population mean must be more than 72​, since the probability is so low.

Step-by-step explanation:

We are given that a geyser has a mean time between eruptions of 72 minutes.

Also, the interval of time between the eruptions be normally distributed with a standard deviation of 23 minutes.

(a) Let X = <u><em>the interval of time between the eruptions</em></u>

So, X ~ N(\mu=72, \sigma^{2} =23^{2})

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

Now, the probability that a randomly selected time interval between eruptions is longer than 82 ​minutes is given by = P(X > 82 min)

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The above probability is calculated by looking at the value of x = 0.43 in the z table which has an area of 0.6664.

(b) Let \bar X = <u><em>sample mean time between the eruptions</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between eruptions has a mean longer than 82 ​minutes is given by = P(\bar X > 82 min)

       P(\bar X > 82 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{82-72}{\frac{23}{\sqrt{13} } } ) = P(Z > 1.57) = 1 - P(Z \leq 1.57)

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The above probability is calculated by looking at the value of x = 1.57 in the z table which has an area of 0.9418.

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                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

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           n = sample of time intervals = 34

Now, the probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 ​minutes is given by = P(\bar X > 82 min)

       P(\bar X > 82 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{82-72}{\frac{23}{\sqrt{34} } } ) = P(Z > 2.54) = 1 - P(Z \leq 2.54)

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The above probability is calculated by looking at the value of x = 2.54 in the z table which has an area of 0.9945.

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(e) If a random sample of 34-time intervals between eruptions has a mean longer than 82 ​minutes, then we conclude that the population mean must be more than 72​, since the probability is so low.

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