Answer:
The equation of the circle having centre (-5,-9) and radius r=3 is
x² + 10 x +y² + 14 y + 65 = 0
Step-by-step explanation:
<u>Explanation</u>:-
From graph The centre of the circle C( -5 , -7)
The radius of the circle r = 3cm from the centre of the given circle
The equation of the circle
From graph the centre ( h, k) = ( -5,-7) and r = 3cm
x² + 10 x +25 +y² + 14 y + 49 =9
x² + 10 x +25 +y² + 14 y + 49 -9=0
The equation of the circle
x² + 10 x +y² + 14 y + 65 = 0
Answer:
The 95% confidence interval for the proportion of parents that are satisfied with their children's education is (0.4118, 0.4618). 0.5 is not part of the confidence interval, so this represents evidence that parents' attitudes toward the quality of education have changed.
Step-by-step explanation:
We have to see if 50% = 0.5 is part of the confidence interval.
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of .
For this problem, we have that:
95% confidence level
So , z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:
The upper limit of this interval is:
The 95% confidence interval for the proportion of parents that are satisfied with their children's education is (0.4118, 0.4618). 0.5 is not part of the confidence interval, so this represents evidence that parents' attitudes toward the quality of education have changed.
Answer:
The probability that there are 3 or less errors in 100 pages is 0.648.
Step-by-step explanation:
In the information supplied in the question it is mentioned that the errors in a textbook follow a Poisson distribution.
For the given Poisson distribution the mean is p = 0.03 errors per page.
We have to find the probability that there are three or less errors in n = 100 pages.
Let us denote the number of errors in the book by the variable x.
Since there are on an average 0.03 errors per page we can say that
the expected value is, = E(x)
= n × p
= 100 × 0.03
= 3
Therefore the we find the probability that there are 3 or less errors on the page as
P( X ≤ 3) = P(X = 0) + P(X = 1) + P(X=2) + P(X=3)
Using the formula for Poisson distribution for P(x = X ) =
Therefore P( X ≤ 3) =
= 0.05 + 0.15 + 0.224 + 0.224
= 0.648
The probability that there are 3 or less errors in 100 pages is 0.648.