A differential equation and various initial conditions are specified. Sketch the graphs of the solutions satisfying these initia
l conditions. For each exercise, put all graphs on one pair of axes.
a. dy/dt=3y(y-2); y(0)=1; y(-2)= -1; y(0)=3; y(0)=2.
b. dy/dt=cos y; y(0)=0; y(-1)=1; y(0)= -pi/2; y(0)=pi.
c. dv/dt= -v2-2v-2; v(0)=0; v(1)=1; v(0)=1
a. dy/dt=3y(y-2); y(0)=1; y(-2)= -1; y(0)=3; y(0)=2.
b. dy/dt=cos y; y(0)=0; y(-1)=1; y(0)= -pi/2; y(0)=pi.
c. dv/dt= -v2-2v-2; v(0)=0; v(1)=1; v(0)=1
1 answer:
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Answer:
A. Value of t = -1.7974 .
B. If the t-score that was calculated was between 1.7 and 1.9, we will not be able to reject the null hypothesis.
C. No, we can't reject the null hypothesis with 99% confidence.
Step-by-step explanation:
We are given the population mean income of students, = $1500.
Let, Null Hypothesis, :
= $1500
Alternate Hypothesis, :
< $1500
A. The one-tailed t statistics used here is ;
follows
where, xbar = sample mean = $1445
s = sample standard deviation = $530
n = sample size = 300
Test Statistics = follows
Value of t = -1.7974
<em>Now, at 5% level of significance the t table will give a critical value of -1.9696. Since our test statistics is greater than critical value as -1.7974 > -1.9696 so we have insufficient evidence to reject null hypothesis and conclude that population mean is $1500 and sample mean is less than the population mean</em>.
B. If the t-score that was calculated above was between 1.7 and 1.9 then also we haven't rejected the null hypothesis as then also our test statistics doesn't lie in the rejection region as 1.7 to 1.9 is greater than -1.9696. And we conclude that sample and the population mean are not same.
C. <em>99% Confidence Interval for </em> =
=
= [ 1384.731 , 1505.269 ]
Since, $1500 lies in this 99% confidence interval so still we can't reject null hypothesis.