What are the first 5 multiples of 150
1 answer:
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Differentiating the function
... g(x) = 5^(1+x)
we get
... g'(x) = ln(5)·5^(1+x)
Then the linear approximation near x=0 is
... y = g'(0)(x - 0) + g(0)
... y = 5·ln(5)·x + 5
With numbers filled in, this is
... y ≈ 8.047x + 5 . . . . . linear approximation to g(x)
Using this to find approximate values for 5^0.95 and 5^1.1, we can fill in x=-0.05 and x=0.1 to get
... 5^0.95 ≈ 8.047·(-0.05) +5 ≈ 4.598 . . . . approximation to 5^0.95
... 5^1.1 ≈ 8.047·0.1 +5 ≈ 5.805 . . . . approximation to 5^1.1
Answer:
Step-by-step explanation:
Given that the Acme Company manufactures widgets, which have a mean of 60 ounces and a standard deviation of 7 ounces
We know that 95% of the area lie between -2 and 2 std deviations from the mean.
i.e. Probability for lying in the middle of 95%
Z score
Between 46 and 74 oz.
b) Between 12 and 57
convert into Z score
P(-6.86<z<-0.43)
=0.5-0.1664=0.3336
c) X<30 gives Z<-4.83
i.e. P(X<30) =0.00