Answer:
There is enough evidence to say that the true average heat output of persons with the syndrmoe differs from the true average heat output of non-sufferers.
Step-by-step explanation:
We have to perform a hypothesis test on the difference between means.
The null and alternative hypothesis are:
μ1: mean heat output for subjects with the syndrome.
μ2: mean heat output for non-sufferers.
We will use a significance level of 0.05.
The difference between sample means is:
The standard error is
The t-statistic is
The degrees of freedom are
The critical value for a left tailed test at a significance level of 0.05 and 16 degrees of freedom is t=-1.746.
The t-statistic is below the critical value, so it lies in the rejection region.
The null hypothesis is rejected.
There is enough evidence to say that the true average heat output of persons with the syndrmoe differs from the true average heat output of non-sufferers.
Answer:
- check below for explanation.
explanation:
➢ Given coordinates:
- A( 1, 3 )
- B( 1, 6 )
- C( 4, 6 )
- D( 5, 2 )
➢ after a dilation with a scale factor of 1/2:
- A( 1, 3 ) ÷ 2 ☛ A'(0.5,1.5)
- B(1, 6 ) ÷ 2 ☛ B'(0.5, 3)
- C( 4, 6) ÷ 2 ☛ C'( 2, 3 )
- D( 5, 2) ÷ 2 ☛ D'( 2.5, 1)
➢ Plot the after dilation coordinates on the graph:
Answer:
c. infinitely many solutions
Step-by-step explanation:
To rearrange the equation given, we need to take note the rules in adding, subtracting, multiplying and dividing terms. We rearrange the equation as follows:
<span>((a+b/c)(d-e)=f) 1/(d-e)
</span>(a+b/c = f/(d-e))c
a+b = fc/(d-e)
a = [fc/(d-e)] - b
Hope this answers the question.
The probability of getting 3 or more who were involved in a car accident last year is 0.126.
Given 9% of the drivers were involved in a car accident last year.
We have to find the probability of getting 3 or more who were involved in a car accident last year if 14 were selected randomly.
We have to use binomial theorem which is as under:
n
where p is the probability an r is the number of trials.
Probability that 3 or more involved in a car accident last year if 14 are randomly selected=1-[P(X=0)+P(X=1)+P(X=2)]
=1-{
}
=1-{1*0.2670+14!/13!*0.9*0.29+14!/2!12!*0.0081*0.2358}
=1-{0.2670+0.3654+0.2358}
=1-0.8682
=0.1318
Among the options given the nearest is 0.126.
Hence the probability that 3 or more are involved in the accident is 0.126.
Learn more about probability at brainly.com/question/24756209
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