(a) Average time to get to school
Average time (minutes) = Summation of the two means = mean time to walk to bus stop + mean time for the bust to get to school = 8+20 = 28 minutes
(b) Standard deviation of the whole trip to school
Standard deviation for the whole trip = Sqrt (Summation of variances)
Variance = Standard deviation ^2
Therefore,
Standard deviation for the whole trip = Sqrt (2^2+4^2) = Sqrt (20) = 4.47 minutes
(c) Probability that it will take more than 30 minutes to get to school
P(x>30) = 1-P(x=30)
Z(x=30) = (mean-30)/SD = (28-30)/4.47 ≈ -0.45
Now, P(x=30) = P(Z=-0.45) = 0.3264
Therefore,
P(X>30) = 1-P(X=30) = 1-0.3264 = 0.6736 = 67.36%
With actual average time to walk to the bus stop being 10 minutes;
(d) Average time to get to school
Actual average time to get to school = 10+20 = 30 minutes
(e) Standard deviation to get to school
Actual standard deviation = Previous standard deviation = 4.47 minutes. This is due to the fact that there are no changes with individual standard deviations.
(f) Probability that it will take more than 30 minutes to get to school
Z(x=30) = (mean - 30)/Sd = (30-30)/4.47 = 0/4.47 = 0
From Z table, P(x=30) = 0.5
And therefore, P(x>30) = 1- P(X=30) = 1- P(Z=0.0) = 1-0.5 = 0.5 = 50%
Answer:
96
Step-by-step explanation:
commom difference = d
a₆ + a₇ = 16
a₅ + a₈ = (a₆ - d) + (a₇ + d) = a₆ + a₇ = 16
a₄ + a₉ = (a₆ - 2d) + (a₇ + 2d) = a₆ + a₇ = 16
Similarly,
a₃ + a₁₀ = 16, a₂ + a₁₁ = 16, a₁ + a₁₂ = 16
so
a₁ + a₂ + ... + a₁₁ + a₁₂ = 6 x 16 = 96
10.5 hours
let t be time then distance travelled by both is
Mr Burges ⇔ d = 50t ( distance = speed × time )
Mrs Burges ⇒ d = 70(t - 3)
At the point of overtaking both will have travelled the same distance
equating their distance and solving for t
70(t - 3) = 50t
70t - 210 = 50t ( subtract 50t from both sides )
20t - 210 = 0 ( add 210 to both sides )
20t = 210 ( divide both sides by 20 )
t = 10.5 hours
Whats the question exactly??
Answer:
False, since x values cannot have more than one y value.
Step-by-step explanation:
