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%.
There will be 42 in each row with 2 left over
To solve this problem, let us first assign some
variables. Let us say that:
x = pigs
y = chickens
z = ducks
From the problem statement, we can formulate the
following equations:
1. y + z = 30 --->
only chicken and ducks have feathers
2. 4 x + 2 y + 2 z = 120 --->
pig has 4 feet, while chicken and duck has 2 each
3. 2 x + 2 y + 2 z = 90 --->
each animal has 2 eyes only
Rewriting equation 1 in terms of y:
y = 30 – z
Plugging this in equation 2:
4 x + 2 (30 – z) + 2 z = 120
4 x + 60 – 2z + 2z = 120
4 x = 120 – 60
4 x = 60
x = 15
From the given choices, only one choice has 15 pigs. Therefore
the answers are:
She has 15 pigs, 12 chickens, and 18 ducks.
54% of 630 = 0.54 * 630 or approximately <u>340</u> of the 630 were planning to vote for the candidate.
51% of 1010 = 0.51 * 1010 or approximately <u>515</u> of the 1010 were planning to vote for the candidate.
This means that the voter support for this candidate increased.
