Then I got 2e=-4
e=-2
e=-2
Five and a half cubed
1 answer:
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Answer: 1.25
Step-by-step explanation:
Given: A college-entrance exam is designed so that scores are normally distributed with a mean = 500 and a standard deviation
= 100.
A z-score measures how many standard deviations a given measurement deviates from the mean.
Let Y be a random variable that denotes the scores in the exam.
Formula for z-score =
Z-score =
⇒ Z-score =
⇒Z-score =1.25
Therefore , the required z-score = 1.25
Answer: (a) e ^ -3x (b)e^-3x
Step-by-step explanation:
I suggest the equation is:
d/dx[integral (e^-3t) dt
First we integrate e^-3tdt
Integral(e ^ -3t dt) as shown in attachment and then we differentiate the result as shown in the attachment.
(b) to differentiate the integral let x = t, and substitute into the expression.
Therefore dx = dt
Hence, d/dx[integral (e ^-3x dx)] = e^-3x
