Answer:
1. If it is true that 60 percent of the trees in a forested region are classified as softwood, 0.015 is the probability of obtaining a population proportion greater than 0.6.
Step-by-step explanation:
Hello!
The historical information indicates that 60% of the forest trees are classified as softwood.
A botanist thinks that the proportion might be greater than 60%, so he tested his belief obtaining:
H₀: p = 0.60
H₁: p > 0.60
p-value: 0.015
You need to interpret this p-value. Little reminder:
The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis. It represents the % of size n samples from a population with proportion p=p₀, which will produce a measure that provides evidence as (or stronger) than the current sample that p is not equal to p₀.
The correct answer is:
1. If it is true that 60 percent of the trees in a forested region are classified as softwood, 0.015 is the probability of obtaining a population proportion greater than 0.6.
I hope this helps!
Answer:
f'(x) = 2[3tan²(x)sec²(x) - 10csc⁴(x)cot(x)]
Step-by-step explanation:
f' of tan(x) = sec²(x)
f' of csc(x) = -csc(x)cot(x)
General Power Rule: uⁿ = xuⁿ⁻¹ · u'
Step 1: Write equation
2tan³(x) + 5csc⁴(x)
Step 2: Rewrite
2(tan(x))³ + 5(csc(x))⁴
Step 3: Find derivative
d/dx 2(tan(x))³ + 5(csc(x))⁴
- General Power Rule: 2 · 3(tan(x))² · sec²(x) + 5 · 4(csc(x))³ · -csc(x)cot(x)
- Multiply: 6(tan(x))²sec²(x) - 20(csc(x))³csc(x)cot(x)
- Simplify: 6tan²(x)sec²(x) - 20csc⁴(x)cot(x)
- Factor: 2[3tan²(x)sec²(x) - 10csc⁴(x)cot(x)]
Answer:
20x-40
Step-by-step explanation:
Answer:
6, 12, 18,
Step-by-step explanation:
u didn't said the numbers but here is the explanation
6÷2=3
6÷3=2
12÷3=4
12÷2=6
18÷2=9
18÷3=6
Answer:
C. 50
Step-by-step explanation:
I would say it looks more like 60 and 50 is the closest