To find the center and radius of the circle represented by the inequality [tex]\displaystyle \sf x^{2} +y^{2} -12y-12\leq 0[/tex], we can complete the square for the y terms.
The inequality can be rewritten as:
[tex]\displaystyle \sf x^{2} +( y^{2} -12y) -12\leq 0[/tex]
To complete the square for the y terms, we need to add and subtract [tex]\displaystyle \sf ( 12/2) ^{2} =36[/tex] inside the parentheses:
[tex]\displaystyle \sf x^{2} +( y^{2} -12y+36) -36-12\leq 0[/tex]
Simplifying, we have:
[tex]\displaystyle \sf x^{2} +( y-6)^{2} -48\leq 0[/tex]
Now we can rewrite the inequality in the standard form of a circle equation:
[tex]\displaystyle \sf ( x-h)^{2} +( y-k)^{2} \leq r^{2}[/tex]
Comparing this with the obtained equation, we can identify the center and radius of the circle:
Center: [tex]\displaystyle \sf ( h,k)=( 0,6)[/tex]
Radius: [tex]\displaystyle \sf r=\sqrt{48}[/tex]
Therefore, the center of the circle is at [tex]\displaystyle \sf ( 0,6)[/tex], and its radius is [tex]\displaystyle \sf \sqrt{48}[/tex].
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♥️ [tex]\large{\underline{\textcolor{red}{\mathcal{SUMIT\:\:ROY\:\:(:\:\:}}}}[/tex]
The darkness of the print is measured quantitatively using an index. If the index is greater than or
equal to 2.0 then the darkness is acceptable. Anything less than 2.0 means the print is too light and
not acceptable. Assume that the machines print at an average darkness of 2.2 with a standard
deviation of 0.20.
(a) What percentage of printing jobs will be acceptable? (4)
(b) If the mean cannot be adjusted, but the standard deviation can, what must be the new standard
deviation such that a minimum of 95% of jobs will be acceptable?
84.13% of the printing jobs will be acceptable.
The new standard deviation required to achieve a minimum of 95% of jobs acceptable is 0.121.
The darkness of the print is measured quantitatively using an index. If the index is greater than or equal to 2.0 then the darkness is acceptable. Anything less than 2.0 means the print is too light and not acceptable. The machines print at an average darkness of 2.2 with a standard deviation of 0.20.
The mean of the darkness of the print is µ = 2.2 and the standard deviation is σ = 0.20.Therefore, the z-score can be calculated as; `z = (x - µ) / σ`.The index required for acceptable prints is 2.0. Thus, the percentage of prints that are acceptable can be calculated as follows;P(X ≥ 2.0) = P((X - µ)/σ ≥ (2.0 - 2.2) / 0.20)P(Z ≥ -1) = 1 - P(Z < -1)Using the standard normal table, P(Z < -1) = 0.1587P(Z ≥ -1) = 1 - 0.1587= 0.8413.
To find the new standard deviation, we can use the z-score formula.z = (x - µ) / σz = (2.0 - 2.2) / σz = -1Therefore, P(X ≥ 2.0) = 0.95P(Z ≥ -1) = 0.95P(Z < -1) = 0.05Using the standard normal table, the z-score value of -1.645 corresponds to a cumulative probability of 0.05. Hence,z = (2.0 - 2.2) / σ = -1.645σ = (2.0 - 2.2) / -1.645= 0.121.
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