46+19+54= 119, so the answer is 119
I think it will be The seconds one so B
Answer:
![x=\frac{1}{4}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B1%7D%7B4%7D)
Step-by-step explanation:
We can use some logarithmic rules to solve this easily.
<em>Note: Ln means
</em>
<em />
Now, lets start with the equation:
![ln(2x) + ln(2) = 0\\ln(2x) = -ln(2)](https://tex.z-dn.net/?f=ln%282x%29%20%2B%20ln%282%29%20%3D%200%5C%5Cln%282x%29%20%3D%20-ln%282%29)
Writing left side with logarithmic base e, we have:
![Log_{e}(2x) = -ln(2)](https://tex.z-dn.net/?f=Log_%7Be%7D%282x%29%20%3D%20-ln%282%29)
We can now use the property shown below to make this into exponential form:
![Log_{a}b=x\\means\\a^x=b](https://tex.z-dn.net/?f=Log_%7Ba%7Db%3Dx%5C%5Cmeans%5C%5Ca%5Ex%3Db)
So, we write:
![Log_{e}(2x) = -ln(2)\\e^{-ln(2)}=2x](https://tex.z-dn.net/?f=Log_%7Be%7D%282x%29%20%3D%20-ln%282%29%5C%5Ce%5E%7B-ln%282%29%7D%3D2x)
We recognize another property of exponentials:
![a^{bc}=(a^{b})^{c}](https://tex.z-dn.net/?f=a%5E%7Bbc%7D%3D%28a%5E%7Bb%7D%29%5E%7Bc%7D)
So, we write:
![e^{-ln(2)}=2x\\(e^{ln(2)})^{-1}=2x](https://tex.z-dn.net/?f=e%5E%7B-ln%282%29%7D%3D2x%5C%5C%28e%5E%7Bln%282%29%7D%29%5E%7B-1%7D%3D2x)
Also, another property of natural logarithms is:
![e^{(ln(a))}=a](https://tex.z-dn.net/?f=e%5E%7B%28ln%28a%29%29%7D%3Da)
Now, we simplify:
![(e^{ln(2)})^{-1}=2x\\(2)^{-1}=2x\\\frac{1}{2}=2x\\x=\frac{\frac{1}{2}}{2}\\x=\frac{1}{4}](https://tex.z-dn.net/?f=%28e%5E%7Bln%282%29%7D%29%5E%7B-1%7D%3D2x%5C%5C%282%29%5E%7B-1%7D%3D2x%5C%5C%5Cfrac%7B1%7D%7B2%7D%3D2x%5C%5Cx%3D%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7D%7D%7B2%7D%5C%5Cx%3D%5Cfrac%7B1%7D%7B4%7D)
This is the answer.
Answer:
Margin of Error = 5.4088 ;
Confidence interval = (30.1 ; 40.9)
Interval estimate are almost the same
Step-by-step explanation:
Given that :
Population standard deviation, σ = 9.3
Sample size, n = 8
Xbar = 35.5
Confidence level = 90%
The confidence interval:
Xbar ± Margin of error
Margin of Error = Zcritical * σ/sqrt(n)
Zcritical at 90% = 1.645
Margin of Error = 1.645 * 9.3/sqrt(8) = 5.4088
Confidence interval :
Xbar ± Margin of error
35.5 ± 5.4088
Lower boundary = (35.5 - 5.4088) = 30.0912 = 30.1
Upper boundary = (35.5 + 5.4088) = 40.9088 = 40.9
(30.1 ; 40.9)
T distribution =. (30.5 ; 40.5)
Normal distribution = (30.1, 40.9)
Answer:
Cuanto dinero voy a pagar 24 $
Cuanto recibiré de cambio 76 $
Step-by-step explanation:
Mis compras son:
1 lápiz 3,50 $
2 plumas 7,90 $
1Libreta 12,60 $
Lo que quiere decir que compre:
3,50 + 7,90 + 12,60 = 24 $
Si cancelo mis compras con un billete de 100 $ me tienen que dar vuelto por:
100 $ - 24 $ = 76 $