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zaharov [31]
2 years ago
15

PLEASE HELP ASAP 25 PTS

Mathematics
1 answer:
stepan [7]2 years ago
6 0

Answer:

2.810  log3(21.903)

Step-by-step explanation:

log5(92) = 2.810

To change bases

logb(a) = logc(a) / logc(b)

 where c is the new base and b is the old base

log5 (92) = log 3(92)/log3(5)

             

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aleksley [76]
The answer would be -0.5.
4 0
3 years ago
Read 2 more answers
Ken has a pole that is 26 inches long. He wishes to cut it into two pieces so that one piece will be 8 inches
bazaltina [42]

Answer:

im too immature for this lol...9. the answer is 9 inches

Step-by-step explanation:

9+9=18

18+8=26

4 0
1 year ago
Please help!<br><br> (ignore the answer I chose)
Andrei [34K]
The numerator factors as (t-8)(t+4), so the whole thing is (t-8)(t+4)/(t-8). Now we can cancel the t-8 AS LONG as t isn't equal to 8, otherwise, we are canceling a zero from both numerator and denominator which is invalid. What is left is t+4.

So your choice was right.
5 0
3 years ago
Choose the best description of the curve of y = a sin x for a &lt; 0, starting at the origin and moving in the positive x direct
Sedaia [141]

Answer: Zero, minimum, Zero, Maximum, Zero

Step-by-step explanation:

That’s the answer in Edge

3 0
2 years ago
In a recent​ year, an author wrote 169 checks. Use the Poisson distribution to find the probability​ that, on a randomly selecte
grigory [225]
We should first calculate the average number of checks he wrote per day.  To do that, divide 169 by 365 (the number of days in a year) and you get (rounded) 0.463.  This will be λ in our Poisson distribution.  Our formula is
P(X=k)= \frac{ \lambda ^{k}-e^{-\lambda} }{k!}.  We want to evaluate this formula for X≥1, so first we must evaluate our case at k=0.  
P(X=0)= \frac{0.463 ^{0}-e ^{-0.463} }{0!} \\ = \frac{1-e ^{-0.463} }{1} =0.3706
To find P(X≥1), we find 1-P(X<1).  Since the author cannot write a negative number of checks, this means we are finding 1-P(X=0).  Therefore we have 1-0.3706=0.6294.
There is a 63% chance that the author will write a check on any given day in the year.<em />
8 0
2 years ago
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