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Rina8888
1 month ago
14

A sample of 75 information systems managers had an average hourly income of $40.75 with a standard deviation of $7.00. The 95% c

onfidence interval for the average hourly wage (in $) of all information system managers is
a. 37.54 to 43.96.
b. 38.61 to 42.89.
c. 39.14 to 42.36.
d. 39.40 to 42.10.
Mathematics
1 answer:
Svet_ta [12.7K]1 month ago
6 0

Answer: c. 39.14 to 42.36

Step-by-step explanation:

We aim to find a 95% confidence interval for the average hourly wage (in $) of all information system managers.

Sample size, n = 75

Mean, u = $40.75

Standard deviation, s = $7.00

For a 95% confidence level, the associated z value is 1.96 as sourced from the normal distribution table.

The formula we will implement is:

Confidence interval

= mean +/- z × standard deviation/√n

This results in:

40.75 +/- 1.96 × 7/√75

= 40.75 ± 1.96 × 0.82

= 40.75 + 1.6072

The lower limit of the confidence interval becomes 40.75 - 1.6072 =39.14

The upper limit is therefore 40.75 + 1.6072 =42.36

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Initially, we need to determine how fast he skis in a minute without considering any speed increase.
To do that, we'll divide the total distance by the time.
960 divided by 5 equals 192.
Therefore, his speed is 192 meters per second.
Now, let's add 20 to this figure.
192 plus 20 equals 212.
Now, to calculate how far he can travel in 10 minutes, we multiply 212 by 10.
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1 month ago
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Alonso brings \$21$21dollar sign, 21 to the market to buy eggs and avocados. He gets eggs that cost \$2.50$2.50dollar sign, 2, p
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Answer:

B ≤ 11 bags

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Alonso starts with $21. After purchasing eggs for $2.50, he has $21 - $2.50 left = $18.50

Thus, the remaining money for avocados is $18.50.

Each 3 bags of avocados cost $5, making the price for 1 bag $(5/3)

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24 days ago
At an ocean-side nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water t
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Answer:

(a1) The chance that the temperature rises by less than 20°C is 0.667.

(a2) The probability of the temperature increase falling between 20°C and 22°C is 0.133.

(b) The likelihood that the temperature increase could be hazardous at any moment is 0.467.

(c) The anticipated value of the temperature rise is 17.5°C.

Step-by-step explanation:

Let X denote the temperature increase.

The random variable X is distributed uniformly over the interval [10°C, 25°C].

The probability density function for X is shown here:

f(X)=\left \{ {{\frac{1}{25-10}=\frac{1}{15};\ x\in [10, 25]} \atop {0;\ otherwise}} \right.

(a1)

The probability of the temperature increase being under 20°C can be calculated as follows:

P(X

Consequently, the chance that the temperature increase will be below 20°C is 0.667.

(a2)

The probability of the temperature rise being in the range from 20°C to 22°C is computed as follows:

P(20

This leads to the probability of the temperature increase being between 20°C and 22°C being 0.133.

(b)

To find the probability that the increase in temperature could be dangerous, we calculate:

P(X>18)=\int\limits^{25}_{18}{\frac{1}{15}}\, dx\\=\frac{1}{15}\int\limits^{25}_{18}{dx}\,\\=\frac{1}{15}[x]^{25}_{18}=\frac{1}{15}[25-18]=\frac{7}{15}\\=0.467

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The expected value of the uniform random variable X is determined as follows:

E(X)=\frac{1}{2}[10+25]=\frac{35}{2}=17.5

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Step-by-step explanation:

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