Answer:
an apparent solution that must be rejected because it does not satisfy the original equation.
Step-by-step explanation:
The functions are illustrations of composite functions.
<em>The soil temperature at 2:00pm is 67</em>
The given parameters are:
---- the function for sun intensity
-- the function for temperature
At 2:00pm, the value of h (number of hours) is:
Substitute 8 for h in , to calculate the sun intensity
Substitute 8/9 for I in , to calculate the temperature of the soil
Approximate
Hence, the soil temperature at 2:00pm is 67
Read more about composite functions at:
brainly.com/question/20379727
Answer:
4u³
Step-by-step explanation:
2/3u × 4/2u² = 4/3u³
4/3u³ × 3/1 = 4u³
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%.
