Answer: option d) (1/16) (42 - x)
Explanation:
1) You forgot to include the table.
This is the table:
Rate time distance
(mi / min) (min) (mi)
up the hill 1/12 x (1/12)x
down the hill 1/16 42-x ?
2) You also forgot to include the set of answerchoices. This is it:
a) x
b) -x
c) 42 - x
d) (1/16) (42-x)
3) Solution:
The missing value is the distance of the trip down the hill.
The distance is the rate times the time, i.e.:
distance = rate * time
the rate is given as 1/16, and the time is given as 42 - x, so the distance is:
distance = (1/16) (42 - x), which is the answer.
Answer:
Yes,there is a significant association shell weight and the widths of the opercula
Step-by-step explanation:
Using a correlation Coefficient calculator :
Given the data above :
The Coefficient of correlation(r) obtained is :
0.7632
Obtaining the test statistic :
T = r² / √(1 - r²) / (n - 2)
T = 0.7632² / √(1 - 0.7632²) / (10 - 2)
T = 0.58247424 / 0.2284528
Test statistic = 2.550
The Pvalue from r score , N = 10
Pvalue(0.7632, 10) = 0.01022
α = 0.05
If Pvalue < α ; reject H0
Pvalue < α ; We conclude that there is a significant association shell weights and the widths of the opercula
Practice, practice, practice. And ask questions.
Sqrt(2) = 1.41
so it would be between 1 and 2
Answer is A
It could be square, rectangle, parallelogram, rhombus/diamond but, trapezoid could not be.
