Use the pythagorean theorem. You should come out with C. Hope this helps!
I didn't get the same answers.
In this problem, 25 is your constant (the number of baby hats you already started out with).
26 will be affected by your variable, d, the number of days. With each day that passes, 26 more hats will be knit, so the expression 26d can be used.
Set h equal to the constant + hats made per day to create an expression that you can use to solve for h:
h = 26d + 25
Now, just plug the numbers in for d to get h:
When d = 2, h = 26d + 25 = 26(2) + 25 = 77
When d = 4, h = 26d + 25 = 26(4) + 25 = 129
When d = 7, h = 26d + 25 = 26(7) + 25 = 207
When d = 9, h = 26d + 25 = 26(9) + 25 = 259
Your answers should be:
77
129
207
259
Answer:
I don't know what you think about it is not going to be a great day of school and I don't know what you think about it is not going to be a great day of school
Answer:
I gotchu the answer is 47
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Answer:
a) P=0.535
b) P=0.204
c) P=0.286
Step-by-step explanation:
The exponential distribution is expressed as

In this example, λ=1/8=0.125 min⁻¹.
a) The probability of having to wait more than 5 minutes

b) The probability of having to wait between 10 and 20 minutes

c) The exponential distribution is memory-less, so it is independent of past events.
If you have waited 5 minutes, the probability of waiting more than 15 minutes in total is the same as the probability of waiting 15-5=10 minutes.
