The resistance at ice point is 5 Ohms and at steam point is 5.23 Ohms. What will be the resistance at 345.65֠C (a) 5.795 Ohms (b) 5.895 Ohms (c) 5.980 Ohms (d) 5.765 Ohms
Answers
Answer:
Answer:
The new mean of new observations is
\displaystyle{\boxed{\red{\sf\:\dfrac{m\:+\:xy}{x}\:}}}
x
m+xy
Step-by-step-explanation:
Let the number of observations be n.
And the observations be a₁, a₂, a₃, a₄,....,aₙ
We have given that,
The mean of certain number of observations is m.
We know that,
\displaystyle{\boxed{\pink{\sf\:Mean\:=\:\dfrac{Sum\:of\:observations}{No.\:of\:observations}}}}
Mean=
No.ofobservations
Sumofobservations
\displaystyle{\implies\sf\:m\:=\:\dfrac{a_1\:+\:a_2\:+\:a_3\:+\:a_4\:+\:\cdots\:+\:a_n}{n}}⟹m=
n
a
1
+a
2
+a
3
+a
4
+⋯+a
n
From the given condition,
Each observation is divided by x ( x ≠ 0 ) and increased by y.
\displaystyle{\therefore\:\sf\:New\:observations\:=\:\dfrac{a_1}{x}\:+\:y\:,\:\dfrac{a_2}{x}\:+\:y\:,\:\dfrac{a_3}{x}\:+\:y\:,\:\cdots\:,\:\dfrac{a_n}{x}\:+\:y}∴Newobservations=
x
a
1
+y,
x
a
2
+y,
x
a
3
+y,⋯,
x
a
n
+y
We have to find the new mean of new observations.
\displaystyle{\boxed{\blue{\:\sf\:New\:mean\:=\:\dfrac{Sum\:of\:new\:observations}{No.\:of\:observations\:}}}}
Newmean=
No.ofobservations
Sumofnewobservations
\displaystyle{\implies\sf\:New\:mean\:=\:\dfrac{\dfrac{a_1}{x}\:+\:y\:+\:\dfrac{a_2}{x}\:+\:y\:+\:\dfrac{a_3}{x}\:+\:y\:+\:\dfrac{a_4}{x}\:+\:y\:+\:\cdots\:+\:\dfrac{a_n}{x}\:+\:y}{n}}⟹Newmean=
n
x
a
1
+y+
x
a
2
+y+
x
a
3
+y+
x
a
4
+y+⋯+
x
a
n
+y
\displaystyle{\implies\sf\:New\:mean\:=\:\dfrac{\dfrac{1}{x}\:(\:a_1\:+\:a_2\:+\:a_3\:+\:a_4\:+\:\cdots\:+\:a_n\:)\:+\:n\:y}{n}}⟹Newmean=
n
x
1
(a
1
+a
2
+a
3
+a
4
+⋯+a
n
)+ny
\displaystyle{\implies\sf\:New\:mean\:=\:\dfrac{a_1\:+\:a_2\:+\:a_3\:+\:a_4\:+\:\cdots\:+\:a_n}{x\:n}\:+\:\dfrac{\cancel{n}\:y}{\cancel{n}}}⟹Newmean=
xn
a
1
+a
2
+a
3
+a
4
+⋯+a
n
+
n
n
y
\displaystyle{\implies\sf\:New\:mean\:=\:\underbrace{\pink{\sf\:\dfrac{a_1\:+\:a_2\:+\:a_3\:+\:a_4\:+\:\cdots\:+\:a_n}{n}}}_{m}\:\times\:\dfrac{1}{x}\:+\:y}⟹Newmean=
m
n
a
1
+a
2
+a
3
+a
4
+⋯+a
n
×
x
1
+y
\displaystyle{\implies\sf\:New\:mean\:=\:m\:\times\:\dfrac{1}{x}\:+\:y}⟹Newmean=m×
x
1
+y
\displaystyle{\implies\sf\:New\:mean\:=\:\dfrac{m}{x}\:+\:y}⟹Newmean=
x
m
+y
\displaystyle{\implies\:\underline{\boxed{\red{\sf\:New\:mean\:=\:\dfrac{m\:+\:xy}{x}\:}}}}⟹
Newmean=
x
m+xy
∴ The new mean of new observations is
\displaystyle{\boxed{\red{\sf\:\dfrac{m\:+\:xy}{x}\:}}}
x
m+xy
Answer:
The new mean of new observations is
\displaystyle{\boxed{\red{\sf\:\dfrac{m\:+\:xy}{x}\:}}}
x
m+xy
Step-by-step-explanation:
Let the number of observations be n.
And the observations be a₁, a₂, a₃, a₄,....,aₙ
We have given that,
The mean of certain number of observations is m.
We know that,
\displaystyle{\boxed{\pink{\sf\:Mean\:=\:\dfrac{Sum\:of\:observations}{No.\:of\:observations}}}}
Mean=
No.ofobservations
Sumofobservations
\displaystyle{\implies\sf\:m\:=\:\dfrac{a_1\:+\:a_2\:+\:a_3\:+\:a_4\:+\:\cdots\:+\:a_n}{n}}⟹m=
n
a
1
+a
2
+a
3
+a
4
+⋯+a
n
From the given condition,
Each observation is divided by x ( x ≠ 0 ) and increased by y.
\displaystyle{\therefore\:\sf\:New\:observations\:=\:\dfrac{a_1}{x}\:+\:y\:,\:\dfrac{a_2}{x}\:+\:y\:,\:\dfrac{a_3}{x}\:+\:y\:,\:\cdots\:,\:\dfrac{a_n}{x}\:+\:y}∴Newobservations=
x
a
1
+y,
x
a
2
+y,
x
a
3
+y,⋯,
x
a
n
+y
We have to find the new mean of new observations.
\displaystyle{\boxed{\blue{\:\sf\:New\:mean\:=\:\dfrac{Sum\:of\:new\:observations}{No.\:of\:observations\:}}}}
Newmean=
No.ofobservations
Sumofnewobservations
\displaystyle{\implies\sf\:New\:mean\:=\:\dfrac{\dfrac{a_1}{x}\:+\:y\:+\:\dfrac{a_2}{x}\:+\:y\:+\:\dfrac{a_3}{x}\:+\:y\:+\:\dfrac{a_4}{x}\:+\:y\:+\:\cdots\:+\:\dfrac{a_n}{x}\:+\:y}{n}}⟹Newmean=
n
x
a
1
+y+
x
a
2
+y+
x
a
3
+y+
x
a
4
+y+⋯+
x
a
n
+y
\displaystyle{\implies\sf\:New\:mean\:=\:\dfrac{\dfrac{1}{x}\:(\:a_1\:+\:a_2\:+\:a_3\:+\:a_4\:+\:\cdots\:+\:a_n\:)\:+\:n\:y}{n}}⟹Newmean=
n
x
1
(a
1
+a
2
+a
3
+a
4
+⋯+a
n
)+ny
\displaystyle{\implies\sf\:New\:mean\:=\:\dfrac{a_1\:+\:a_2\:+\:a_3\:+\:a_4\:+\:\cdots\:+\:a_n}{x\:n}\:+\:\dfrac{\cancel{n}\:y}{\cancel{n}}}⟹Newmean=
xn
a
1
+a
2
+a
3
+a
4
+⋯+a
n
+
n
n
y
\displaystyle{\implies\sf\:New\:mean\:=\:\underbrace{\pink{\sf\:\dfrac{a_1\:+\:a_2\:+\:a_3\:+\:a_4\:+\:\cdots\:+\:a_n}{n}}}_{m}\:\times\:\dfrac{1}{x}\:+\:y}⟹Newmean=
m
n
a
1
+a
2
+a
3
+a
4
+⋯+a
n
×
x
1
+y
\displaystyle{\implies\sf\:New\:mean\:=\:m\:\times\:\dfrac{1}{x}\:+\:y}⟹Newmean=m×
x
1
+y
\displaystyle{\implies\sf\:New\:mean\:=\:\dfrac{m}{x}\:+\:y}⟹Newmean=
x
m
+y
\displaystyle{\implies\:\underline{\boxed{\red{\sf\:New\:mean\:=\:\dfrac{m\:+\:xy}{x}\:}}}}⟹
Newmean=
x
m+xy
∴ The new mean of new observations is
\displaystyle{\boxed{\red{\sf\:\dfrac{m\:+\:xy}{x}\:}}}
x
m+xy