Traingle BUG has measurements as shown below.
1 answer:
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First, let's use the given information to determine the function's amplitude, midline, and period.
Then, we should determine whether to use a sine or a cosine function, based on the point where x=0.
Finally, we should determine the parameters of the function's formula by considering all the above.
Determining the amplitude, midline, and period
The midline intersection is at y=5 so this is the midline.
The maximum point is 1 unit above the midline, so the amplitude is 1.
The maximum point is π units to the right of the midline intersection, so the period is 4 * π.
Determining the type of function to use
Since the graph intersects its midline at x=0, we should use thesine function and not the cosine function.
This means there's no horizontal shift, so the function is of the form -
a sin(bx)+d
Since the midline intersection at x=0 is followed by a maximumpoint, we know that a > 0.
The amplitude is 1, so |a| = 1. Since a >0 we can conclude that a=1.
The midline is y=5, so d=5.
The period is 4π so b = 2π / 4π = 1/2 simplified.
= Solution
Then, we should determine whether to use a sine or a cosine function, based on the point where x=0.
Finally, we should determine the parameters of the function's formula by considering all the above.
Determining the amplitude, midline, and period
The midline intersection is at y=5 so this is the midline.
The maximum point is 1 unit above the midline, so the amplitude is 1.
The maximum point is π units to the right of the midline intersection, so the period is 4 * π.
Determining the type of function to use
Since the graph intersects its midline at x=0, we should use thesine function and not the cosine function.
This means there's no horizontal shift, so the function is of the form -
a sin(bx)+d
Since the midline intersection at x=0 is followed by a maximumpoint, we know that a > 0.
The amplitude is 1, so |a| = 1. Since a >0 we can conclude that a=1.
The midline is y=5, so d=5.
The period is 4π so b = 2π / 4π = 1/2 simplified.
Solution :
Let and
represents the proportions of the seeds which germinate among the seeds planted in the soil containing
and
mushroom compost by weight respectively.
To test the null hypothesis against the alternate hypothesis
.
Let denotes the respective sample proportions and the
represents the sample size respectively.
The test statistic can be written as :
which under follows the standard normal distribution.
We reject at
level of significance, if the P-value
or if
Now, the value of the test statistics = -1.368928
The critical value =
P-value =
= 0.171335
Since the p-value > 0.05 and , so we fail to reject
at
level of significance.
Hence we conclude that the two population proportion are not significantly different.
Conclusion :
There is not sufficient evidence to conclude that the of the seeds that
with the percent of the
in the soil.