The whole equation simplified to one answer is 5
Answer:
Given A triangle ABC in which
∠C =90°,∠A=20° and CD ⊥ AB.
In Δ ABC
⇒∠A + ∠B +∠C=180° [ Angle sum property of triangle]
⇒20° + ∠B + 90°=180°
⇒∠B+110° =180°
∠B =180° -110°
∠B = 70°
In Δ B DC
∠BDC =90°,∠B =70°,∠BC D=?
∠BDC +,∠B+∠BC D=180°[ angle sum property of triangle]
90° + 70°+∠BC D =180°
∠BC D=180°- 160°
∠BC D = 20°
In Δ AC D
∠A=20°, ∠ADC=90°,∠AC D=?
∠A + ∠ADC +∠AC D=180° [angle sum property of triangle]
20°+90°+∠AC D=180°
110° +∠AC D=180°
∠AC D=180°-110°
∠AC D=70°
So solution are, ∠AC D=70°,∠ BC D=20°,∠DB C=70°
f(x)g(x) = 9x² + 10x - 16
<u>Step-by-step explanation:</u>
Step 1:
Given f(x) = 3x - 4 × 2 + 6x and g(x) = 5x + 2 × 2 - 3x
f(x) = 9x - 8
g(x) = 2x + 4
f(x)g(x) = (9x - 8)(2x + 4)
= 18x² + 36x - 16x -32
= 18x² + 20x - 32
= 9x² + 10x - 16
Step-by-step explanation:
Keep quit when they a teaching you and stay with participanting people/friends
Answer:
e) The mean of the sampling distribution of sample mean is always the same as that of X, the distribution from which the sample is taken.
Step-by-step explanation:
The central limit theorem states that
"Given a population with a finite mean μ and a finite non-zero variance σ2, the sampling distribution of the mean approaches a normal distribution with a mean of μ and a variance of σ2/N as N, the sample size, increases."
This means that as the sample size increases, the sample mean of the sampling distribution of means approaches the population mean. This does not state that the sample mean will always be the same as the population mean.
