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1 year ago

Given that g(x)=3(x)^2-5(x)+1 g(1-t)

Mathematics

1 answer:

Vinvika [58]1 year ago

7 0

Answer: g(1-t)=3t^2-t-1

Step-by-step explanation:

g(x)=3(x)^2-5(x)+1

g(1-t)=3(1-t)^2-5(1-t)+1

g(1-t)=3(1-t)(1-t)-5-(-5t)+1

g(1-t)=3(1-t-t+t^2)-5+5t+1

g(1-t)=3(1-2t+t^2)-5+5t+1

g(1-t)=3-6t+3t^2-5+5t+1

g(1-t)=3-5+1-6t+5t+3t^2

g(1-t)=-1-t+3t^2

g(1-t)=3t^2-t-1

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Step-by-step explanation:

Step 1: Convert to Improper fractions

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Using the normal distribution we can find z value with the mean and standard deviation as follows. x is the value we want to know its probability

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