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GREYUIT [131]
3 years ago
15

The duration of routine operations at chicago general hospital has approximately a normal distribution with an average of 130 mi

nutes and a standard deviation of 12 minutes. what proportion of operations last at least 120 minutes
Mathematics
1 answer:
vredina [299]3 years ago
5 0

Answer:

79.77%

Step-by-step explanation:

P(X≥120) = normalcdf(120,1e99,130,12) = 0.7976716754 ≈ 0.7977 = 79.77%

Therefore, about 79.77% of operations last at least 120 minutes

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Solve -1/2 w -3/5 = 1/5w
Alika [10]

Answer:

-6/7.

Step-by-step explanation:

-1/2 w -3/5 = 1/5w

-1/2 w - 1/5 w = 3/5

Multiply through by 10:

-5w - 2w = 6

-7w = 6

w = -6/7.

4 0
3 years ago
Multiply 12 by the sum of 8 and "t"
Katen [24]
Mathematically, that is 12(8+t). Distribute the 12, and you get 96 + 12t, or 12t + 96.
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4 years ago
The amount of time all students in a very large undergraduate statistics course take to complete an examination is distributed c
Anestetic [448]

Answer:

a) The mean is \mu = 60

b) The standard deviation is \sigma = 9

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

The probability a student selected at random takes at least 55.50 minutes to complete the examination equals 0.6915.

This means that when X = 55.5, Z has a pvalue of 1 - 0.6915 = 0.3085. This means that when X = 55.5, Z = -0.5

So

Z = \frac{X - \mu}{\sigma}

-0.5 = \frac{55.5 - \mu}{\sigma}

-0.5\sigma = 55.5 - \mu

\mu = 55.5 + 0.5\sigma

The probability a student selected at random takes no more than 71.52 minutes to complete the examination equals 0.8997.

This means that when X = 71.52, Z has a pvalue of 0.8997. This means that when X = 71.52, Z = 1.28

So

Z = \frac{X - \mu}{\sigma}

1.28 = \frac{71.52 - \mu}{\sigma}

1.28\sigma = 71.52 - \mu

\mu = 71.52 - 1.28\sigma

Since we also have that \mu = 55.5 + 0.5\sigma

55.5 + 0.5\sigma = 71.52 - 1.28\sigma

1.78\sigma = 71.52 - 55.5

\sigma = \frac{(71.52 - 55.5)}{1.78}

\sigma = 9

\mu = 55.5 + 0.5\sigma = 55.5 + 0.5*9 = 55.5 + 4.5 = 60

Question

The mean is \mu = 60

The standard deviation is \sigma = 9

6 0
3 years ago
What's the answer please and thank you ​
topjm [15]

Where's the question??

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3 years ago
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The ratios of side lengths in a 30-60-90 triangle are 1 : √3 : 2.

The longer leg is ...
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