State the converse Please helps asap
1 answer:
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Answer:
. hole or removable discontinuity at x=2
Step-by-step explanation:
Well generally if you want the simplest form, you factor each the denominator and numerator and then see if you can cancel any of the factors out (because they're in the denominator and numerator)
So let's start by factoring the first equation:
Now let's find what ac is (it's just c since a=1...)
List factors of -6
.
Now we have to look for two numbers that add up to -5. It's a bit obvious here since there isn't many factors, but it's -2 and -3, and they're both negative since 6 is positive, and -5 is negative...
So using these two factors we get
Ok now let's factor the second equation:
Multiply a and c
List factors of 12:
.
Factors that add up to -7 and multiply to 12:
Rewrite equation:
Group terms:
Factor out GCF:
Rewrite:
Now let's write out the equation using these factors:
.
Here we can factor out the x-2 and the simplified form is:
So we can "technically" define f(2) using the most simplified form, but it's removable discontinuity, so it has a hole as x=2. since it makes (x-2) equal to 0 (2-2) = 0.
Answer:
a. Null hypothesis:
Alternative hypothesis:
b.
c.
So with the p value obtained and using the significance level given we have
so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).
Step-by-step explanation:
a. State and label the null and alternative hypotheses.
The system of hypothesis on this case are:
Null hypothesis:
Alternative hypothesis:
Or equivalently:
Null hypothesis:
Alternative hypothesis:
Our notation on this case :
represent the sample size for group 1
represent the sample size for group 2
represent the sample mean for the group 1
represent the sample mean for the group 2
represent the sample standard deviation for group 1
represent the sample standard deviation for group 2
b. State the value of the test statistic.
And the statistic is given by this formula:
Where t follows a t distribution with degrees of freedom. If we replace the values given we have:
Now we can calculate the degrees of freedom given by:
c. Find either the critical value(s) and draw a picture of the critical region(s) or find the P-value for this test. Indicate which method you are using: ( CIRCLE ONE: Critical value / P-value )
Method used: P value
And now we can calculate the p value using the altenative hypothesis, since it's a right tail test the p value is given by:
So with the p value obtained and using the significance level given we have
so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).