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nikdorinn [45]
2 years ago
8

The mean per capita income is 19,292 dollars per annum with a variance of 540,225. What is the probability that the sample mean

would be less than 19269 dollars if a sample of 499 persons is randomly selected? Round your answer to four decimal places.
Mathematics
1 answer:
Snezhnost [94]2 years ago
5 0

Answer:

The probability is 0.2423.

Step-by-step explanation:

Given mean per capita = 19292 dollars

Given the variance = 540225

Now find the probability that the sample mean will be less than 19269 dollar when the sample is 499.

Below is the calculation:

\bar{X} \sim N(\mu =19292, \ \sigma  = \frac{\sqrt{540225}}{\sqrt{499}}) \\\bar{X} \sim N(\mu =19292, \ \sigma  = 32.90) \\\text{therefore the probability is:} \\P (\bar{X}< 19269) \\\text{Convert it to standard normal variable.} \\P(Z< \frac{19269-19292}{32.90}) \\P(Z< - 0.6990) \\\text{Now getting the probability from standard normal table}\\P(Z< -0.6990) = 0.2423

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Find the dimensions of a rectangle with area 512 m2 whose perimeter is as small as possible. (If both values are the same number
Masja [62]

Answer:

<h2>√512 by √512 </h2>

Step-by-step explanation:

Length the length and breadth of the rectangle be x and y.

Area of the rectangle A = Length * breadth

Perimeter P = 2(Length + Breadth)

A = xy and P = 2(x+y)

If the area of the rectangle is 512m², then 512 = xy

x = 512/y

Substituting x = 512/y into the formula for calculating the perimeter;

P = 2(512/y + y)

P = 1024/y + 2y

To get the value of y, we will set dP/dy to zero and solve.

dP/dy = -1024y⁻² + 2

-1024y⁻² + 2 = 0

-1024y⁻² = -2

512y⁻² = 1

y⁻² = 1/512

1/y² = 1/512

y²  = 512

y = √512 m

On testing for minimum, we must know that the perimeter is at the minimum when y = √512

From xy = 512

x(√512) = 512

x = 512/√512

On rationalizing, x = 512/√512 * √512 /√512

x = 512√512 /512

x = √512 m

Hence, the dimensions of a rectangle is √512 m  by √512 m

5 0
2 years ago
Will likes two brands of healthy breakfast cereal. In Superfiber cereal, there are 5 grams of fiber in one cup. In Fiber Oats ce
Katen [24]

Let x  <u>represent</u> the <u>number of cups</u> of Superfiber Will ate this week and let y  represent the number cups of Fiber Oats he ate this week.

1. If in Superfiber cereal, there are 5 grams of fiber in one cup, then in x cups there are 5\cdot x=5x grams of fiber.

2. If in Fiber Oats cereal, there are 4 grams of fiber in one cup, then in y cups there are 4\cdot y=4y grams of fiber.

3. The total amount of fiber Will ate is 5x+4y grams.

4. This week Will ate at least 30 grams of fiber. Then

5x+4y\ge 30.

Answer: correct choice is A.


7 0
2 years ago
Read 2 more answers
The commute time to work in the U.S. Has a bell shaped distribution with a population mean of 24.4 minutes and a population stan
Schach [20]

Answer:

Q1 a) 0.9545

b) 0.02275

c) 0.97725

Q2 z is approximately equal to -1.466

Step-by-step explanation:

Q1 The given information are;

The mean time to commute to work = 24.4 minutes

The standard deviation = 6.5 minutes

a) The z-score for 11.4 is given as follows;

Z=\dfrac{x-\mu }{\sigma }

Where;

x = Observed value 11.4

μ = The mean = 24.4 minutes

σ = The standard deviation = 6.5 minutes

Z=\dfrac{11.4-24.4 }{6.5 } = -2

The z-score for 37.4 is given as follows;

Z=\dfrac{37.4-24.4 }{6.5 } = 2

-2 < z < 2, which gives, from the z-score table;

The probability of commute time to be between 11.4 minutes and 37.4 minutes =  0.97725 - 0.02275 = 0.9545

b) From the z-score table, the probability that the commute time to be less than 11.4 minutes = The probability at z = -2 = 0.02275

c) From the z-score table, the probability that the commute time to be greater than 37.4 minutes = The probability at z = 2 = 0.97725

Q2 The the z-score that corresponds to a commute time of 15 minutes is given as follows;

Z=\dfrac{x-\mu }{\sigma }

Z=\dfrac{15-24.4 }{6.5 } = -\dfrac{94}{65} \approx -1.466

7 0
2 years ago
If 8 identical blackboards are to be divided among 4 schools,how many divisions are possible? How many, if each school mustrecei
MAXImum [283]

Answer:

There are 165 ways to distribute the blackboards between the schools. If at least 1 blackboard goes to each school, then we only have 35 ways.

Step-by-step explanation:

Essentially, this is a problem of balls and sticks. The 8 identical blackboards can be represented as 8 balls, and you assign them to each school by using 3 sticks. Basically each school receives an amount of blackboards equivalent to the amount of balls between 2 sticks: The first school gets all the balls before the first stick, the second school gets all the balls between stick 1 and stick 2, the third school gets the balls between sticks 2 and 3 and the last school gets all remaining balls.

 The problem reduces to take 11 consecutive spots which we will use to localize the balls and the sticks and select 3 places to put the sticks. The amount of ways to do this is {11 \choose 3} = 165 . As a result, we have 165 ways to distribute the blackboards.

If each school needs at least 1 blackboard you can give 1 blackbooard to each of them first and distribute the remaining 4 the same way we did before. This time there will be 4 balls and 3 sticks, so we have to put 3 sticks in 7 spaces (if a school takes what it is between 2 sticks that doesnt have balls between, then that school only gets the first blackboard we assigned to it previously). The amount of ways to localize the sticks is {7 \choose 3} = 35. Thus, there are only 35 ways to distribute the blackboards in this case.

4 0
2 years ago
Hiromi sells 12 T-shirts each week at a price of $13.00. Past sales have shown that for every $0.25 decrease in price, 4 more T-
Lerok [7]
(13-0.25x4)(12+4x4)=2,464
(13-0.25x5)(12+4x5)=3,845
(13-0.25x6)(12+4x6)=2,845
so your answer is 3,845 because if Hirome sells them for $11.75 he makes $3845 if Hirome sold them for $11.50 he makes $2845.80 and if he sold them for $12.00 he would make $2464.00

4 0
2 years ago
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