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Answer:
Step-by-step explanation:
1. Null hypothesis: u <= 0.784
Alternative hypothesis: u > 0.784
2. Find the test statistics: z using the one sample proportion test. First we have to find the standard deviation
Using the formula
sd = √[{P (1-P)}/n]
Where P = 0.84 and n = 750
sd =√[{0.84( 1- 0.84)/750]}
sd=√(0.84 (0.16) /750)
SD =√(0.1344/750)
sd = √0.0001792
sd = 0.013
Then using this we can find z
z = (p - P) / sd
z = (0.84-0.784) / 0.013
z =(0.056/0.013)
z = 4.3077
3. Find the p value and use it to make conclusions...
The p value at 0.02 level of significance for a one tailed test with 4.3077 as z score and using a p value calculator is 0.000008254.
4. Conclusions: the results is significant at 0.02 level of significance suck that we can conclude that its on-time arrival rate is now higher than 78.4%.
Step-by-step explanation:
sol;
x+1=y...(1)
3y-7=2x....(2)
or, 3(x+1)-72x [from (1)]
or, 3x+3-7=2x
or, 3x-2x=7-3
x=4
now,
putting the value of x in (1)
y=x+1
=4+1
=5
PR and SQ are the diagonal of PQRS.
3/7, 3/5, 1/2 is the answer it starts from the lowest to the lagest
