Answer: 1/2 1 over 2 if you didn’t know that’s fraction form and decimal form it’s 0.5
Answer:
12
Step-by-step explanation:
A compound interest equation is represented as A = P(1 + r/n)^nt, where n=the number of times per year the interest is compounded. In the given equation, n=12.
Answer:
The value of the single logarithm is -11 ⇒ answer B
Step-by-step explanation:
* Lets revise the rule of the logarithmic functions
# ㏒ a + ㏒ b = ㏒ ab
# ㏒ a - ㏒ b = ㏒ a/b
# ㏒ a^n = n ㏒ a
# ㏒ 1 = 0
* Now lets solve the problem
∵ ![log_{b}(\frac{A^{5}C^{2}}{D^{6}})](https://tex.z-dn.net/?f=log_%7Bb%7D%28%5Cfrac%7BA%5E%7B5%7DC%5E%7B2%7D%7D%7BD%5E%7B6%7D%7D%29)
- Change the single logarithm to an expression by change the
multiplication to addition and the division to subtraction
∵ ![log_{b}A^{5}=5log_{b}A](https://tex.z-dn.net/?f=log_%7Bb%7DA%5E%7B5%7D%3D5log_%7Bb%7DA)
∵ ![log_{b}C^{2}=2log_{b}C](https://tex.z-dn.net/?f=log_%7Bb%7DC%5E%7B2%7D%3D2log_%7Bb%7DC)
∵ ![log_{b}D^{6}=6log_{b}D](https://tex.z-dn.net/?f=log_%7Bb%7DD%5E%7B6%7D%3D6log_%7Bb%7DD)
∴ The single logarithm = ![5log_{b}A+2log_{b}C-6log_{b}D](https://tex.z-dn.net/?f=5log_%7Bb%7DA%2B2log_%7Bb%7DC-6log_%7Bb%7DD)
* Now lets substitute the values
∵ ![log_{b}A=3;===log_{b}C=2;===log_{b}D=5](https://tex.z-dn.net/?f=log_%7Bb%7DA%3D3%3B%3D%3D%3Dlog_%7Bb%7DC%3D2%3B%3D%3D%3Dlog_%7Bb%7DD%3D5)
- Substitute the values into the expression
∴ The value = 5(3) + 2(2) - 6(5) = 15 + 4 - 30 = -11
* The value of the single logarithm is -11
Answer:
Ellipses (special case is called a circle), hyperbolas, parabolas.
Step-by-step explanation:
These are all conic sections.
A conic section is defined by the cross section of a plane and a double-napped cone. There are other special cases called degenerate conics, which are lines and points (occurs when the equation does not follow the usual pattern, however, these are not considered main conics). The main types of conics are: ellipses, hyperbolas, and parabolas.
The illustration below gives more insight into the question.
I hope this helps.
Using a calculator with the binompdf and binomcdf features, I can calculate these values. My calculator is a TI-83 plus, and the features are found under the 2nd, Vars keys (Scroll up or down until you see them).
If "exactly" is to be found, use binompdf:
binompdf(number of trials, probability of success, exactly number)
ANSWER for exactly 3: binompdf(8, 0.5, 3) = 0.21875 = 21.875%
If "at least" is to be found, use binomcdf:
binomcdf(number of trials, probability of success, at least number - 1)
ANSWER for at least 6: binomcdf(8, 1/2, 5) ≈ 0.8555 ≈ 85.55%
If "at most" is to be found, use binomcdf:
binomcdf(number of trials, probability of success, at most number)
ANSWER for at most 3: binomcdf(8, 0.5, 3) ≈ 0.3633 ≈ 36.33%