23.6*2.5=59 inches of average rainfall for that month
A) Composite function that represents how many flowers Iris can expect to bloom over a certain number of weeks is f[s(w)] = 50w + 25.
B) The unit of measurement for the composite function is flowers.
C) Number of the flowers for 30 weeks will be 1525.
<h3>What is a composite function?</h3>
A function is said to be a composite function when a function is written in another function. The composite function that represents the number of flowers is f[s(w)] = 50w + 25. and the number of flowers for 30 weeks is 1525.
Part A: Write a composite function that represents how many flowers Iris can expect to bloom over a certain number of weeks.
From the given data we will find the function for the number of flowers with time.
f(s) = 2s + 25
We have s(w) = 25w
f[(s(w)]=2s(w) + 25
f[(s(w)] = 2 x ( 25w ) +25
f[s(w)] = 50w + 25.
Part B: What are the units of measurement for the composite function in Part A
The expression f[s(w)] = 50w + 25 will give the number of the flowers blooming over a number of the weeks so the unit of measurement will be flowers.
Part C: Evaluate the composite function in Part A for 30 weeks.
The expression f[s(w)] = 50w + 25 will be used to find the number of flowers blooming in 30 weeks put the value w = 30 to get the number of the flowers.
f[s(w)] = 50w + 25.
f[s(w)] = (50 x 30) + 25.
f[s(w)] = 1525 flowers.
Therefore the composite function is f[s(w)] = 50w + 25. unit will be flowers and the number of flowers in 30 weeks will be 1525.
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The hypothesis test shows that we reject the null hypothesis and there is sufficient evidence to support the claim that the return rate is less than 20%
<h3>What is the claim that the return rate is less than 20% by using a statistical hypothesis method?</h3>
The claim that the return rate is less than 20% is p < 0.2. From the given information, we can compute our null hypothesis and alternative hypothesis as:
Given that:
Sample size (n) = 6965
Sample proportion 
The test statistics for this data can be computed as:
z = -2.73
From the hypothesis testing, since the p < alternative hypothesis, then our test is a left-tailed test(one-tailed.
Hence, the p-value for the test statistics can be computed as:
P-value = P(Z ≤ z)
P-value = P(Z ≤ - 2.73)
By using the Excel function =NORMDIST (-2.73)
P-value = 0.00317
P-value ≅ 0.003
Therefore, we can conclude that since P-value is less than the significance level at ∝ = 0.01, we reject the null hypothesis and there is sufficient evidence to support the claim that the return rate is less than 20%
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Answer:
x ≥ 1
Step-by-step explanation:
→ As it is a shaded circle you know it ≤ or ≥
x ≥ 1
Answer:
Im not sure how to be honest.
Step-by-step explanation:
Im not sure how to be honest.
