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Romashka-Z-Leto [24]
3 years ago
15

For some value of z, the value of the cumulative standardized normal distribution is 0.8340. the value of z is

Mathematics
1 answer:
Aleksandr-060686 [28]3 years ago
6 0

For some value of z, the value of the cumulative standardized normal distribution is 0.8340. the value of z is

Answer: We are required to find the value of z corresponding to probability 0.8340.

i.e., P(Z

We can find the value of z using the standard normal table.

Using the standard normal table, we have:

z(0.8340)=0.97

Therefore, for the value of z = 0.97, cumulative standardized normal distribution is 0.8340

Attached here standard normal table for your reference.



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\fbox{ \fbox { \sf{Distance  =  \sf{15 \: units}}}}

{ \fbox { \fbox { \sf{Midpoint = { \sf{ \: (12 \: , \: 10.5)}}}}}}

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\star{ \:  \sf { \: Let \: the \: points \: be \: A \: and \: B}}

\star { \sf{Let \: A(6 \:, 6) \: be \: (x1 ,\: y1) \: and \: B(18 ,\: 15) \: be \: (x2 \:, y2)}}

\underline{ \underline{ \tt{Finding \: the \: distance}}}

\boxed{ \sf{Distance =  \sqrt{ {(x2 - x1)}^{2}  +  {(y2 - y1)}^{2} } }}

\mapsto{ \sf{Distance =  \sqrt{ {(18 - 6)}^{2}  +  {(15 - 6)}^{2} } }}

\mapsto{ \sf{Distance =  \sqrt{ {(12)}^{2}  +  {(9)}^{2} } }}

\mapsto{ \sf{Distance =  \sqrt{144 + 81}}}

\mapsto{ \sf{Distance =  \sqrt{225} }}

\mapsto{ \sf{Distance =  \sqrt{ {15}^{2} } }}

\mapsto{  \boxed{\sf{Distance = 15 \: units}}}

\underline{ \underline {\tt{Finding \: the \: Midpoint}}}

\boxed{ \sf{Midpoint = ( \frac{x1 + x2}{2}  \: , \:  \frac{y1 + y2}{2} )}}

\mapsto{ \sf{Midpoint = ( \frac{6 + 18}{2}  \: , \:  \frac{6 + 15}{2} )}}

\mapsto{ \sf{Midpoint = ( \frac{24}{2}  \: , \:  \frac{21}{2} )}}

\mapsto{ \boxed{ \sf{Midpoint = (12 \: , \: 10.5)}}}

Hope I helped!

Best regards! :D

~\sf{TheAnimeGirl}

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