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poizon [28]
3 years ago
13

Jeff lives 12 miles east of Stan. Jeff lives 16 miles north of Wei. What is the shortest distance that Stan and Wei can live fro

m each other?
Mathematics
1 answer:
Greeley [361]3 years ago
7 0
|  \
|     \ 
| 16  \
|         \
|______\
 12

Envision this as a right triangle, with one leg of 12 miles, and the other 16 miles. Solve using the Pythagorean Theorem a²+b²=c²:

12²+16²=c²
400 = c²
20=c...they live at least 20 miles from each other.

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The graph of a function is given. Determine whether the function is increasing, decreasing, or constant on the given interval. (
Alika [10]

Answer: Constant

Step-by-step explanation:

5 0
3 years ago
The angles of a pentagon are x, x − 5 0 , x + 100 , 2x + 150and 2x + 300 . Find all the angles.
miv72 [106K]

Answer:

The interior angles are 70°,65°,80°,155° and 170°

Step-by-step explanation:

step 1

Find the sum of the interior angles of the pentagon

The sum is equal to

S=(n-2)*180°

where

n is the number of sides of polygon

n=5 (pentagon)

substitute

S=(5-2)*180°=540°

step 2

Find the value of x

Sum the given angles and equate to 540

x+(x-5)+(x+10)+(2x+15)+(2x+30)=540°

7x+50=540°

7x=490°

x=70°

step 3

Find all the angles

x=70°

(x-5)=(70-5)=65°

(x+10)=(70+10)=80°

(2x+15)=(2*70+15)=155°

(2x+30)=(2*70+30)=170°

8 0
2 years ago
The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.9 minutes and a standard deviation of 2.9
Eva8 [605]

Answer:

a) 0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) 0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes

c) 0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 8.9 minutes and a standard deviation of 2.9 minutes.

This means that \mu = 8.9, \sigma = 2.9

Sample of 37:

This means that n = 37, s = \frac{2.9}{\sqrt{37}}

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

320/37 = 8.64865

Sample mean below 8.64865, which is the p-value of Z when X = 8.64865. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{8.64865 - 8.9}{\frac{2.9}{\sqrt{37}}}

Z = -0.53

Z = -0.53 has a p-value of 0.2981

0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

275/37 = 7.4324

Sample mean above 7.4324, which is 1 subtracted by the p-value of Z when X = 7.4324. So

Z = \frac{X - \mu}{s}

Z = \frac{7.4324 - 8.9}{\frac{2.9}{\sqrt{37}}}

Z = -3.08

Z = -3.08 has a p-value of 0.001

1 - 0.001 = 0.999

0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Sample mean between 7.4324 minutes and 8.64865 minutes, which is the p-value of Z when X = 8.64865 subtracted by the p-value of Z when X = 7.4324. So

0.2981 - 0.0010 = 0.2971

0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

7 0
2 years ago
How do you write 2/10 as a percentage
slava [35]
It’s 20% because 10/10 would be 100% and 1/10 would be 10%
4 0
3 years ago
The cost of holding a children's birthday party at a rollerskating rink is a function of n, the number of people in the party. T
Bas_tet [7]

Answer: It would cost $150 for a party of 9 people and it would cost $260 for 20 people

Step-by-step explanation:

The number 9 falls in between 0<n<_ 12 so the cost is 150

The number 20 falls in between 12<n<_ 20 so the cost is 260

4 0
2 years ago
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