Answer:
0.09803922 if its 25 dived by 255 but if you wrote it wrong and it's actually 255 divided by 25 it would be 9
Step-by-step explanation:
<h3>
Answer:</h3>
- a_n = -3a_(n-1); a_1 = 2
- a_n = 2·(-3)^(n-1)
<h3>
Step-by-step explanation:</h3>
A) The problem statement tells you it is a geometric sequence, so you know each term is some multiple of the one before. The first terms of the sequence are given, so you know the first term. The common ratio (the multiplier of interest) is the ratio of the second term to the first (or any term to the one before), -6/2 = -3.
So, the recursive definition is ...
... a_1 = 2
... a_n = -3·a_(n-1)
B) The explicit formula is, in general, ...
... a_n = a_1 · r^(n -1)
where r is the common ratio and a_1 is the first term. Filling in the known values, this is ...
... a_n = 2·(-3)^(n-1)
Answer:
g(x) approaches negative infinity
Step-by-step explanation:
g(x) = -x² + 2x + 4
limit at g(x) approaches infinity = -(∞)² + 2(∞) + 4 = -∞
I can't solve for f(x) because you didnt give me f(x)
Area = 1/2 * 10 * 4 = 20 <span>sq. inches
</span><span>B. 20 sq. inches</span>
Answer:
1. E(Y) = 50.54°F
2. SD(Y) = 11.34°F
Step-by-step explanation:
We are given that The daily high temperature X in degrees Celsius in Montreal during April has expected value E(X) = 10.3°C with a standard deviation SD(X) = 3.5°C.
The conversion of X into degrees Fahrenheit Y is Y = (9/5)X + 32.
(1) Y = (9/5)X + 32
E(Y) = E((9/5)X + 32) = E((9/5)X) + E(32)
= (9/5) * E(X) + 32 {
expectation of constant is constant}
= (9/5) * 10.3 + 32 = 50.54
Therefore, E(Y), the expected daily high in Montreal during April in degrees Fahrenheit is 50.54°F .
(2) Y = (9/5)X + 32
SD(Y) = SD((9/5)X + 32) = SD((9/5)X) + SD(32)
= 
* SD(X) + 0 { standard deviation of constant is zero}
= 
* 3.5 = 11.34°F
Therefore, SD(Y), the standard deviation of the daily high temperature in Montreal during April in degrees Fahrenheit is 11.34°F .
