log(x) + log(3) = log(18)
log(x) + 0.477121 = 1.255273
Add -0.477121 to both sides.
log(x) + 0.477121 + −0.477121 = 1.255273 + −0.477121
log(x)+0 = 0.778152
Divide both sides by 1.
log(x)+0/1 = 0.778152/1
log(x)=0.778152
Solve Logarithm
log(x) = 0.778152
10log(x) = 100.778152
x = 100.778152
x = 6.00001
Answer:
Option B.
Step-by-step explanation:
Let as consider the given equation:
It can be written as
![[\because \ln e^a=a]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Cln%20e%5Ea%3Da%5D)
![[\because \ln a^b=b\ln a]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Cln%20a%5Eb%3Db%5Cln%20a%5D)
![[\because \ln \dfrac{a}{b}=\ln a-\ln b]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Cln%20%5Cdfrac%7Ba%7D%7Bb%7D%3D%5Cln%20a-%5Cln%20b%5D)
On comparing both sides, we get
Multiply both sides by 5.
Divide both sides by 2.
Therefore, the correct option is B.
Answer:
(-2,3)
so the a answer is -2 on the x axis
Using it's concept, it is found that the probabilities are given as follows:
a.1/2, 1/9, 5/9.
<h3>What is a probability?</h3>
A probability is given by the <u>number of desired outcomes divided by the number of total outcomes</u>.
In this problem, 90 out of 180 individuals are children, hence the probability is given by:
p = 90/180 = 1/2.
Out of 90 adults, 60 - 50 = 10 preferred cheeseburgers, hence the probability is given by:
p = 10/90 = 1/9.
Of the 180 individuals, 90 are children, and 10 are adults who prefer cheeseburgers, hence the probability is given by:
p = 100/180 = 10/18 = 5/9.
Hence option a is correct.
More can be learned about probabilities at brainly.com/question/14398287
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Answer:
0.4394
Step-by-step explanation:
given that an auto parts store is examining how many items were purchased per transaction at the store as compared to their online website
The data is shown below:
Frequency Frequency*Midpt
Midpt Online Total Online Total
1-3 2 40 147 80 294
4-6 5 60 103 300 515
7-9 8 15 32 120 256
10-12 11 5 18 55 198
120 300 555 1263
Thus we find total items purchased on line = 555
and total items overall purchased = 1263
Hence the probability that a randomly selected item was purchased online
=
