We know that the original width was 5, and that after the enlargement the width is 45. To calculate the scale factor we divide the final width between the original one, then:
Therefore the scale factor is 9.
13u = 26 is it 39, 338 , or 2
2 answers:
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Refer to the diagram shown below.
w = 6 7/8 in = 6.875 in, the width of each device.
d = 3 1/2 in = 3.50 in, the space between teo devices.
The total space needed is
D = 4(w+d) + w
= 5w + 4d
= 5*6875 + 4*3.5
D = 48.375 in or 48 3/8 in
Answer: 48 3/8 inches or 48.375 inches
w = 6 7/8 in = 6.875 in, the width of each device.
d = 3 1/2 in = 3.50 in, the space between teo devices.
The total space needed is
D = 4(w+d) + w
= 5w + 4d
= 5*6875 + 4*3.5
D = 48.375 in or 48 3/8 in
Answer: 48 3/8 inches or 48.375 inches
Answer:
69.50%
Step-by-step explanation:
Given:
Ayrshire cows:
E(X) =μ = 47
SD(X) = σ = 6 Var (X) = 6^2 = 36
Jersey cows:
E(Y) = μ = 43
SD(Y) = σ = 5 Var(Y) = 52 = 25
Properties mean, variance and standard deviation:
E(X +Y) = E(X) E(Y)
V ar(X +Y) = Var(X) + Var(Y)
SD(X +Y) = √Var(X)+Var(Y)
X — Y represents the difference between Ayrshire and Jersey cows.
E(X — Y) = E(X) — E(Y). 47 — 43 = 4
SD(X — Y) = √Var(X)+ Var(Y) =√36+ 25 = √61 = 7.8102
The z-score is the value decreased by the mean, divided by the standard deviation:
z = x-μ /σ = 0-4/ 7.8102 = -0.51
Determine the corresponding probability using table Z in appendix F.
P(X—Y 0) = P(Z > —0.51) = 1—P(Z < —0.51) = 1-0.3050 = 0.6950 = 69.50%