Answer:
p = 0.293
Conclusion:
"There is insufficient evidence to state that the distribution of fish in Green Valley Lake is not the same as the distribution of fish in Echo Lake."
Step-by-step explanation:
------------R/trout--O/trout---bass--catfish---total
GVL - - - 105 - - - 27 - - - - - 35--- 24 - - - 191
EL - - - - 135------- 48-------- 67 - - 43 - - - 293
Total - - 240 - - - 75 - - - - - 102--- 67 - - - 484
Note : GVL = GREEN VALLEY LAKE ; EL = ECHO LAKE.
H0: The distribution of fish caught in green valley is the same as that caught in Echo lake.
H1: The distribution of fish caught in Green Valley lake is different from that caught in Echo valley lake.
Using the chi-squared test statistic calculator ;
χ2 = 3.7269
p-value = 0.2925.
p-value = 0.293
Conclusion:
"There is insufficient evidence to state that the distribution of fish in Green Valley Lake is not the same as the distribution of fish in Echo Lake."
Answer:
The absolute brightness of the Cepheid star after a period of 45 days is -5.95
Step-by-step explanation:
Since the absolute magnitude or brightness of a Cepheid star is related to its period or length of its pulse by
M = –2.78(log P) – 1.35 where M = absolute magnitude and P = period or length of pulse.
From our question, it is given that P = 45 days.
So, M = –2.78(log P) – 1.35
M = –2.78(log 45) – 1.35
M = –2.78(1.6532) – 1.35
M = -4.60 - 1.35
M = -5.95
So, the absolute magnitude or brightness M of a Cepheid star after a period P of 45 days is -5.95
Supplementary angles are two angles that when add equal 180 degrees.
The supplement of an angle is 180 - n.
n = 17(180 - n)
n = 3060 - 17n
n + 17n = 3060
18n = 3060
n = 3060/18
n = 170
so the unknown angle (n) = 170 degrees and its supplement = (180 - 170 = 10)......= 10 degrees.
To solve
82% of x = 119.31
rewrite
x = 119.31/0.82
solve
x <span>≈ 145.5</span>
Hope this helps :)
Old
9.99×55
=549.45
New
10.68×55
=587.4
((10.68÷9.99)−1)×100
=6.9%
