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frez [133]
3 years ago
13

The manager of a computer help desk operation has collected enough data to conclude that the distribution of time per call is no

rmally distributed with a mean equal to 8.21 minutes and a standard deviation of 2.14 minutes. what is the probability that three randomly monitored calls will each be completed in 4 minutes or less?
Mathematics
1 answer:
kvv77 [185]3 years ago
5 0

Answer:

0.0003

Step-by-step explanation:

Mean=μ=8.21

Standard deviation=σ=2.14

We have to find P(3 randomly monitored call completed in 4 min or less).

P(Xbar≤4)=?

μxbar=μ=8.21

σxbar=σ/√n=2.14/√3=1.2355

Z-score associated with xbar=4

Z=[Xbar-μxbar]/σxbar

Z=[4-8.21]/1.2355

Z=-4.21/1.2355

Z=-3.4075

P(Xbar≤4)=P(Z≤-3.41)

P(Xbar≤4)=P(-∞<Z<0)-P(0<Z<-3.41)

P(Xbar≤4)=0.5-0.4997

P(Xbar≤4)=0.0003

Thus, the probability that three randomly monitored calls will each be completed in 4 minutes or less is 0.0003.

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klio [65]

Answer:

4.5/3 or 1.5

Step-by-step explanation:

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(6.5-2)/7-4)=4.5/3... you can further simplify to 1.5.

8 0
3 years ago
PLEASE HELP GIVING MANY POINTS!
NeTakaya

Answer:

  69.01 m

Step-by-step explanation:

The mnemonic SOH CAH TOA reminds you ...

  Tan = Opposite/Adjacent

The tangent function is useful for problems like this. Let the height of the spire be represented by h. The distance (d) across the plaza from the first surveyor satisfies the relation ...

  tan(50°) = (h -1.65)/d

Rearranging to solve for d, we have ...

  d = (h -1.65)/tan(50°)

The distance across the plaza from the second surveyor satisfies the relation ...

  tan(30°) = (101.65 -h)/d

Rearranging this, we have ...

  d = (101.65 -h)/tan(30°)

Equating these expressions for d, we can solve for h.

  (h -1.65)/tan(50°) = (101.65 -h)/tan(30°)

  h(1/tan(50°) +1/tan(30°)) = 101.65/tan(30°) +1.65/tan(50°)

We can divide by the coefficient of h and simplify to get ...

  h = (101.65·tan(50°) +1.65·tan(30°))/(tan(30°) +tan(50°))

  h ≈ 69.0148 ≈ 69.01 . . . . meters

The tip of the spire is 69.01 m above the plaza.

3 0
3 years ago
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nikdorinn [45]
I think it’s the second option
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