You use a line of best fit for a set of data to make a prediction about an unknown value. The correlation coefficient for your d
ata set is 0.214. How confident can you be that your predicted value will be reasonably close to the actual value?
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1 answer:
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Answer:
1) 20.9
2) 896
3) 21
Step-by-step explanation:
1) 5.6÷2^3+(12.75+7.45)
---> 12.75 + 7.45 = 20.2
÷
--> Simplify 2^3 to 8
÷ 8 + 20.2
--> 5.6 ÷ 8 = 0.7
--> Simplify
2) 4^3 × (0.6 +3.6) ÷ 0.3
---> 0.6 + 3.6 = 4.2
4^3 * 4.2 ÷ 0.3
---> 4^3 = 64
64 * 4.2 ÷ 0.3
--> 64 * 4.2 = 268.8
268.8 ÷ 0.3
--> 268.8 ÷ 0.3 = 896
896
3) 2^4 + (2.75 +1.75) ÷ 0.9
--> 2.75 + 1.75 = 4.5
2^4 + 4.5 ÷ 0.9
--> 2^4 = 16
16 + 4.5 ÷ 0.9
--> 4.5 ÷ 0.89 = 5
16 + 5
--> Simplify
= 21
a. The inverse image of f(x) is f⁻¹(x) = ∛(x/3)
b. The inverse image of g(x) is
c. The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9
<h3 /><h3>The domain of T</h3>Since T = {x ∈ R : 0 ≤ [x^2 − 25} ⇒ x² - 25 ≥ 0
⇒ x² ≥ 25
⇒ x ≥ ±5
⇒ -5 ≤ x ≤ 5.
<h3>Inverse image of f(x)</h3>
The inverse image of f(x) is f⁻¹(x) = ∛(x/3)
f : R → R defined by f(x) = 3x³
Let f(x) = y.
So, y = 3x³
Dividing through by 3, we have
y/3 = x³
Taking cube root of both sides, we have
x = ∛(y/3)
Replacing y with x we have
y = ∛(x/3)
Replacing y with f⁻¹(x), we have
So, the inverse image of f(x) is f⁻¹(x) = ∛(x/3)
<h3>Inverse image of g(x)</h3>
The inverse image of g(x) is
g : R+ → R defined by g(x) = ln(x).
Let g(x) = y
y = ln(x)
Taking exponents of both sides, we have
Replacing x with y, we have
Replacing y with g⁻¹(x), we have
So, the inverse image of g(x) is
The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9
h : R → R defined by h(x) = x − 9
Let y = h(x)
y = x - 9
Adding 9 to both sides, we have
y + 9 = x
Replacing x with y, we have
x + 9 = y
Replacing y with h⁻¹(x), we have
So, the inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9
Learn more about inverse image of a function here: