Answer:
Option (C)
Step-by-step explanation:
Domain of a function is represented by the 'x' values on the graph.
Similarly, y-values of a function represents the range.
Therefore, options (B) and (D) will not be the answer as they are representing the function values or y-values.
Given segments starts form x = -2 and ends at x = 2 (Including these points).
Therefore, Domain of the segment will be [-2, 2] Or -2 ≤ x ≤2
Option (C) will be the correct option.
Answer:
B) 1 ≤ ƒ(x) ≤ 6
Step-by-step explanation:
this is the right answer I got it right on usa test prep
Answer:
x = 5°
Step-by-step explanation:
We know that in a triangle, the measure of an exterior angle is equal to the sum of its two remote interior angles, therefore:
7x + 4 + 61 = 20x
7x + 65 = 20x
13x = 65
x = 5°
Answer:
Solution given:
61°+(7x+4)°=20x [ exterior angle is equal to the sum of two opposite interior angle]
65+7x=20x
65=20x-7x
13x=65°
x==5°
value of x=5°
b. 0.869
c. 0.847
d. 0.680
Answer:
c. 0.847
Step-by-step explanation:
From the given information;
Given that n = 500 which is too large, binomial distribution can now be approximated to
where;
∴
From the z table
Thus, our value is closest to the option c which 0.847
Answer:
Step-by-step explanation:
The answer is 56 mph
Answer:
Step-by-step explanation:
We assume you want your model to be ...
p = c·e^(kt)
Filling in (t, p) values of (3, 484) and (5, 1135), we have two equations in the two unknowns:
484 = c·e^(3k)
1135 = c·e^(5k)
Taking logs makes these linear equations:
ln(484) = ln(c) +3k
ln(1135) = ln(c) +5k
Subtracting the first equation from the second, we have ...
ln(1135) -ln(484) = 2k
k = ln(1135/484)/2 ≈ 0.42615
Using that value in the first equation, we find ...
ln(484) = ln(c) +3(ln(1135/484)/2)
ln(c) = ln(484) -(3/2)ln(1135/484)
c = e^(ln(484) -(3/2)ln(1135/484)) ≈ 134.8
The initial number in the culture was 135, and the k-value is about 0.42615.
_____
I prefer to start with the model ...
p = 484·(1135/484)^((t-3)/2)
Then the initial value is that obtained when t=0:
c = 484·(1135/484)^(-3/2) = 134.778 ≈ 135
The value of k the log of the base for exponent t. It is ...
ln((1135/484)^(1/2)) = 0.426152
This starting model matches the given numbers exactly. The transformation to c·e^(kt) requires approximations that make it difficult to match the given numbers.
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For this model, the base of the exponent is the ratio of the two given population values. The exponent is horizontally offset by the number of days for the first count, and scaled by the number of days between counts. The multiplier of the exponential term is the first count. The model can be written directly from the given data, with no computation required.