Answer:
209.005 gms
Step-by-step explanation:
Given that the weights of packets of cookies produced by a certain manufacturer have a Normal distribution with a mean of 202 grams and a standard deviation of 3 grams.
Let X be the weight of packets of cookies produced by manufacturer
X is N(202, 3) gms.
To find the weight that should be stamped on the packet so that only 1% of the packets are underweight
i.e. P(X<c) <0.01
From std normal table we find that z value = 2.335
Corresponding x value = 202+3(2.335)
=209.005 gms.
Answer:
B: 0.01p + 0.05(22 – p) = 0.54
Step-by-step explanation:
B on edge
Answer:
-2 15 ...................
Answer:
For number 3 the answer is -4
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.
Below is the solution:
xdy/dx = 4y
<span>xdy = 4ydx </span>
<span>∫ dy/y =∫ 4dx/x </span>
<span>ln(y) = 4ln(x) + C </span>
<span>y = e^[4ln(x)] e^C </span>
<span>y = e^[ln(x)^4] e^C </span>
<span>y = Cx^4 answer</span>
