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
Answer:
1/000000000
is the standard form of 10 to the power minus 9
Answer:
44
Step-by-step explanation:
1) we know that opposite sides of a parallelogram are congruent so
4x+28=10x+4
2) subtract 4 on both sides
4x+24=10x
3)subtract 4x on both sides
24=6x
4)divide by 6 on both sides
4=x
5)substitute for x
length of RS =10(4)+4
length of RD=44
No, this is not normal
This is a binomial distribution. Use BINS to determine if this is binomial.
B - binary?
I - independent?
N - number of trials
S - probability of success
B - yes, 3 or not a 3
I - yes, past rolls do not impact future rolls
N - 20 trials
S - prob success 1/6
Use binompdf on your calculator to find out the probability. To access, 2nd —> vars —> binompdf
Binompdf(20 (trials), 1/6 (p), 11 (x)) = 8.97x10^-5
The probability to roll a 3 11 times is .0000897. The chances are very low, making this not normal.