Help please I'm to stupid for this
1 answer:
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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
Answer:
Left up x=34
Left down x=87
Right x=17
Step-by-step explanation:
The picture is the working out.
This is an example of conditional probability because we are trying to find the probability of an event occurring GIVEN the occurrence of some other event. There is a formula for this (see image attached).
If we follow this formula, the numerator would be the probability of (A AND B) which in this case is "48% of the class passed BOTH exams." The denominator in the formula would be that "60% of the class passed ONLY THE SECOND exam."
Therefore, P(A and B) = 0.48, which is 48% expressed as a decimal and P(B)= 0.60, which is 60% expressed as a decimal. Then, you can figure out the answer by dividing.
If we follow this formula, the numerator would be the probability of (A AND B) which in this case is "48% of the class passed BOTH exams." The denominator in the formula would be that "60% of the class passed ONLY THE SECOND exam."
Therefore, P(A and B) = 0.48, which is 48% expressed as a decimal and P(B)= 0.60, which is 60% expressed as a decimal. Then, you can figure out the answer by dividing.