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alexira [117]
3 years ago
10

Five hundred randomly selected working adults living in Calgary, Canada were asked how long, in minutes, their typical daily com

mute was (Calgary Herald Traffic Study, Ipsos, September 17, 2005). The resulting sample mean and standard deviation of commute time were 28.5 minutes and 24.2 minutes, respectively. Construct and interpret a 90% confidence interval for the mean commute time of working adult Calgary residents.
Mathematics
1 answer:
Anarel [89]3 years ago
5 0

Answer:

90% confidence interval for the mean commute time of working adult Calgary residents is between a lower limit of 26.72 minutes and an upper limit of 30.28 minutes.

Step-by-step explanation:

Confidence Interval = Mean + or - Error margin (E)

mean = 28.5 minutes

sd = 24.2 minutes

n = 500

degree of freedom = n - 1 = 500 - 1 = 499

confidence level = 90%

t-value corresponding to 499 degrees of freedom and 90% confidence level = 1.64801

E = t × sd/√n = 1.64801 × 24.2/√500 = 1.78

Lower limit = mean - E = 28.5 - 1.78 = 26.72 minutes

Upper limit = mean + E = 28.5 + 1.78 = 30.28 minutes

90% confidence interval is between 26.72 and 30.28 minutes.

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Write the equation that shows 40 is 8 times more than 5
Vlad1618 [11]

Answer:

8X5=40

Step-by-step explanation:

3 0
3 years ago
What is FB? Please Explain the steps to find it.
Troyanec [42]
Check the picture below.

8 0
3 years ago
Find the sum of the first 47 terms of the following series, to the nearest integer.
Leokris [45]

Answer:

The sum of the first 47 terms of the series, 12, 16, 20, ... S₄₈ is 4,888

Step-by-step explanation:

The given series is;

12, 16, 20, ...,

Therefore, the first term of the series is, a = 12

The common difference of series is found as follows;

The difference between subsequent terms, 12 and 16 is  16 - 12 = 4

The difference between subsequent terms, 16, and 20 is  20 - 16 = 4

Therefore, the common difference, d = 4

The series is therefore an arithmetic projection, AP

The sum of the first 'n' terms of an AP, Sₙ, is given as follows;

S_n = \dfrac{n}{2} \cdot \left [2 \cdot a + (n - 1)\cdot d \right ]

(47/2)*(2*12+(47-1)*4)

The sum of the first 47 terms is therefore given as follows;

S_n = \dfrac{47}{2} \cdot \left [2 \times 12 + (47 - 1)\times 4 \right ] = 4,888

The sum of the first 47 terms of the series, 12, 16, 20, ... S₄₈ = 4,888

5 0
2 years ago
5 yrs ago, Nuri was thrice as old as Sonu. 10 yrs later, Nuri will be twice as old Sonu. How old are Nuri n Sonu?​
Papessa [141]

Answer:

Answer will be 50

Step-by-step explanation:

Let us suppose, present age of Nuri be ‘x’ years and present age of Sonu be ‘y’ years.

Now, it is given that five years ago, Nuri was thrice old as Sonu. Hence,

Five years ago,

Nuri’s age = x-5 years

Sonu’s age = y-5 years

And relation between ages can be given as

Nuri’s age = 3×sonu’s age or

x-5 = 3(y-5)

x-5 = 3y-15

x-3y+10 = 0 ………..(i)

Another relation is given in the problem that ten years later, Nuri is twice as old as Sonu.

So, ten years ago,

Nuri’s Age = x+10

Sonu’s Age = y+10

And relation between ages can be written as

x+10 = 2(y+10)

x+10 = 2y+20

x-2y-10 = 0 …………..(ii)

Now we can solve the equation (i) and (ii) to get values of x and ‘y’ or present ages of Nuri and Sonu.

Value of ‘x’ from equation (i) be

x = 3y-10 ……….(iii)

Putting value of ‘x’ from equation (iii) in equation (ii) we get,

3y-10-2y-10 = 0

y = 20

Now, from equation (iii) value of ’x’ can be given as,

x= 3(20)-10

x = 50

Hence, the present ages of Nuri and Sonu are 50 years and 20 years respectively.

7 0
2 years ago
You’re given two side lengths of 3 centimeters and 5 centimeters. Which measurement can you use for the length of the third side
mrs_skeptik [129]

Answer:

You could use a measurement of 4 centimeters.

The valid range is  2  < x < 8  cms.

Step-by-step explanation:

The third side must either be less than 3+5, that is less than 8 cms and  it must be greater than (5-3) = 2 cms. If it is any other length we could not construct a triangle.

8 0
3 years ago
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