Joshua started cycling at 5:15 pm By 8:09 pm, he has covered a distance of 7250 mWhat was Joshua's average speed during that tim
1 answer:
The average speed of Joshua during that time is 2500 m/h.
Explanation:
It is given that Joshua started cycling at 5:15 pm. By 8:09 pm he has covered a distance of 7250 m.
The total time taken by Joshua from 5:15 pm to 8:09 pm is
Dividing we get,
Adding, we have,
Thus, the total time taken by Joshua is
To determine the average speed we use the formula,
where and
Hence, substituting the values we have,
Dividing, we get,
Thus, the average speed of Joshua during that time is 2500 m/h.
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Answer:
The correct options are 1 and 3.
Step-by-step explanation:
From the given box plot it is clear that
Q₁ is 25% of a data.
Median is 50% of a data.
Q₃ is 75% of a data.
34 is minimum value of the data and 46 is median it means 50% of the data values lies between 34 and 46. Therefore option 1 is correct.
42 is first quartile and and 70 is third quartile. it means 50% of the data values lies between 42 and 70. Therefore option 2 is incorrect.
The difference between Minimum value and first quartile, Maximum value and third quartile is less than 1.5×(IQR), therefore it is unlikely to have any outliers in the data.
Hence option 3 is correct.
The interquartile range of the data is
The interquartile range is 28. Therefore option 4 is incorrect.
Range of the data is
The range is 42. Therefore option 5 is incorrect.
Answer:
<em>X ball bearings will be made in 1 hour and 20 minutes</em>
Step-by-step explanation:
<u>Proportions </u>
The proportions give us an important tool to easily solve common problems of any type. We know that:
Company ABC can make X ball bearings in 3 hours.
Company DEF can make X ball bearings in 4 hours.
Company GHI can make X ball bearings in 6 hours.
In one hour, each company can make
ABC: X/3 ball bearings
DEF: X/4 ball bearings
GHI: X/6 ball bearings
Working together, they can make
ball bearings. Operating
If they can make 3/4 of a ball bearing in one hour, then one complete ball bearing will need
hours to complete. It's equivalent to 1 1/3 hours or 1 hour and 20 minutes
X ball bearings will be made in 1 hour and 20 minutes
Answer:
a) There is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.
b) There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.
c) There is a 68.74% 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:
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
.
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
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 pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. This means that .
(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?
We are working with a sample mean of 37 jets. So we have that:
Total time of 320 minutes for 37 jets, so
This probability is the pvalue of Z when . So
has a pvalue of 0.7422. This means that there is a 74.22% 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?
Total time of 275 minutes for 37 jets, so
This probability is subtracted by the pvalue of Z when
has a pvalue of 0.0548.
There is a 1-0.0548 = 0.9452 = 94.52% 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?
Total time of 320 minutes for 37 jets, so
Total time of 275 minutes for 37 jets, so
This probability is the pvalue of Z when subtracted by the pvalue of Z when
.
So:
From a), we have that for , we have
, that has a pvalue of 0.7422.
From b), we have that for , we have
, that has a pvalue of 0.0548.
So there is a 0.7422 - 0.0548 = 0.6874 = 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.
Answer:
(a) 0.7967
(b) 0.6826
(c) 0.3707
(d) 0.9525
(e) 0.1587
Step-by-step explanation:
The random variable <em>X</em> follows a Normal distribution with mean <em>μ</em> = 10 and variance <em>σ</em>² = 36.
(a)
Compute the value of P (X > 5) as follows:
Thus, the value of P (X > 5) is 0.7967.
(b)
Compute the value of P (4 < X < 16) as follows:
Thus, the value of P (4 < X < 16) is 0.6826.
(c)
Compute the value of P (X < 8) as follows:
Thus, the value of P (X < 8) is 0.3707.
(d)
Compute the value of P (X < 20) as follows:
Thus, the value of P (X < 20) is 0.9525.
(e)
Compute the value of P (X > 16) as follows:
Thus, the value of P (X > 16) is 0.1587.
**Use a <em>z</em>-table for the probabilities.