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vesna_86 [32]
3 years ago
5

How to find the shape of distribution ??

Mathematics
1 answer:
Natali5045456 [20]3 years ago
8 0
The shape of the distribution of the time required to get an oil change at a 10 - minute oil - change facility is Unknown. However, records indicate that the mean time is 11. 2 minutes, and the standard deviation is 3. 5 minutes (a) To compute probabilities regarding the sample mean Using the normal model, what size sample would be required? (b) What is the probability that a random sample of n = LO oil changes results in a sample mean time less than 10 minutes?
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Start time 2 p.m. elapsed time blank and time 8:30 p.m.​
Nataly_w [17]

Answer:

Well I guess this is a math problem but the middle time would be: 6 hours and 30 minutes.

Step-by-step explanation:

Adding 6 hours to 2 pm makes 8:00 on the dot at night. Adding 30 minutes to 8:00 would be 8:30. There you have your answer.

3 0
3 years ago
Plzzzzzzzzzzz help i will give brainliest
lara [203]

Answer:

Below.

Step-by-step explanation:

IQR is the same

Number of data points is the same.

Mode - can't tell

Range - different

First quartile - same

Median - different.

3 0
3 years ago
I don't know who needs to hear this but I wanted to hopefully reach out to someone.
Alex787 [66]

Answer:

aw thank you sm <3 !!! needed to hear that

Step-by-step explanation:

5 0
4 years ago
Read 2 more answers
What is the correct answer for this 5+1*10=
Viefleur [7K]
It would equal 15. To get 15, you would first do 1*10= which equals 10. After that, you would add 5 to the 10 then there's your answer. I hoped this helped!
8 0
4 years ago
Read 2 more answers
The 1992 world speed record for a bicycle (human-powered vehicle) was set by Chris Huber. His time through the measured 200 m st
Reil [10]

Answer:

a) 30.726m/s and b) 5.5549s

Step-by-step explanation:

a.) What was Chris Huber’s speed in meters per second(m/s)?

Given the distance and time, the formula to obtain the speed is

v=\frac{d}{t}.

Applying this to our problem we have that

v=\frac{200m}{6.509s}= 30.726m/s.

So, Chris Huber’s speed in meters per second(m/s) was 30.726m/s.

b) What was Whittingham’s time through the 200 m?

In a) we stated that v=\frac{d}{t}. This formula implies that

  1. t=\frac{d}{v}.

First, observer that 19\frac{km}{h} =19,000\frac{m}{h}=\frac{19,000}{3,600}m/s= 5.2777m/s.

Then, Sam Whittingham speed was equal to Chris Huber’s speed plus 5.2777 m/s. So, v=30.726\frac{m}{s} +5.2777\frac{m}{s}= 36.003 m/s.

Then, applying 1) we have that

t=\frac{200m}{36.003m/s}=5.5549s.

So, Sam Whittingham’s time through the 200 m was 5.5549s.

5 0
4 years ago
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