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Musya8 [376]
4 years ago
3

From a population that is not normally distributed and whose standard deviation is not known, a sample of 6 items is selected to

develop an interval estimate for the mean of the population (μ).
a. The normal distribution can be used.
b. The t distribution with 6 degrees of freedom must be used.
c. The sample size must be increased.
d. The t distribution with 5 degrees of freedom must be used.
Mathematics
1 answer:
Bumek [7]4 years ago
4 0

Answer:

d) The t-distribution with 5 degrees of freedom must be used

Step-by-step explanation:

For cases of Normal Distribution where the variance is unknown and the sample size n is smaller than 30, we must use the t-student distribution.

The shape of the curve for t-student is bell-shape (flatter and with wider tails than the bell shape of normal distribution.

Actually, when we deal with t-student distribution we are dealing with a family of curves that will become closer and closer to the bell shape of the normal distribution as the degree of freedom increases. From values of n =30( and bigger),  we can assume that the curve of t-student is the same as for normal distribution

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Equivalent to 6.4 kilograms
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14.1096 lb (pounds)

Step-by-step explanation:


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A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 43 ft/s. Its height
Lina20 [59]

Answer:

Instantaneous Velocity at t = 1 is 20 feet per second

Step-by-step explanation:

We are given he following information in the question:

y(t)=43t-23t^{2}

B) Instantaneous Velocity at t = 1

y(1) = 43(1)-23(1)^2 = 20

A) Formula:

Average velocity =

\displaystyle\frac{\text{Displacement}}{\text{Time}}

1) 0.01

y(1 + 0.01)=y(1.01) = 43(1.01)-23(1.01)^{2} = 19.9677\\\\\text{Average Velocity} = \dfrac{y(1.01)-y(1)}{1.01-1} =\dfrac{19.9677-20}{1.01-1}= \dfrac{-0.0323}{0.01} = -3.230000 \text{feet per second}

2) 0.005 s

y(1 + 0.005)=y(1.005) = 43(1.005)-23(1.005)^{2} = 19.984425\\\\\text{Average Velocity} = \dfrac{y(1.005)-y(1)}{1.005-1} =\dfrac{19.984425-20}{1.005-1} = -3.1150000 \text{feet per second}

3) 0.002 s

y(1 + 0.002)=y(1.002) = 43(1.002)-23(1.002)^{2} = 19.993908\\\\\text{Average Velocity} = \dfrac{y(1.002)-y(1)}{1.002-1} =\dfrac{19.993908-20}{1.002-1} = -3.0460000 \text{feet per second}

4) 0.001 s

y(1 + 0.001)=y(1.001) = 43(1.001)-23(1.001)^{2} = 19.996977\\\\\text{Average Velocity} = \dfrac{y(1.001)-y(1)}{1.001-1} =\frac{19.996977-20}{1.001-1} = -3.0230000 \text{feet per second}

3 0
3 years ago
a truck costing 25000 with a residual value of 5000. the truck's estimated life is 10 years. what is the book value at the end o
sineoko [7]
$25,000 - 5,000 = $20,000
Since the useful life is 10 years or 10% per year, double-declining rate would be 20%

Year 1
$20,000 x .2 = $4000
Book Value = $16,000

Year 2
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3 0
4 years ago
The compound interest on a sum of money in
Yakvenalex [24]

Answer:

The difference between the principal and the compound interest in three years is Rs 17,994

Step-by-step explanation:

The compound interest is given according to the following formula;

C.I. = P \cdot \left ( 1 + \dfrac{r}{n} \right ) ^{n\cdot t} - P

The given amount of the compound after 2 years = Rs 5,460

The given amount of the compound after 4 years = Rs 12,066.60

Therefore, we have;

5,460 = P \cdot \left ( 1 + \dfrac{r}{100} \right ) ^{2} - P...(1)

12,066.60 = P \cdot \left ( 1 + \dfrac{r}{100} \right ) ^{4} - P ...(2)

Dividing equation (2) by (1), we have;

\dfrac{12,066.60}{5,460} = \dfrac{P \cdot \left ( \left ( 1 + \dfrac{r}{100} \right ) ^{4} - 1\right )}{P \cdot \left (\left ( 1 + \dfrac{r}{100} \right ) ^{2} -1 \right ) } =\dfrac{\left ( 1 + \dfrac{r}{100} \right ) ^{4} - 1}{\left ( 1 + \dfrac{r}{100} \right ) ^{2} -1  }

Let \  \left ( 1 + \dfrac{r}{100} \right ) ^{2} = x, we \ get;

\dfrac{12,066.60}{5,460} =\dfrac{\left ( 1 + \dfrac{r}{100} \right ) ^{4} - 1}{\left ( 1 + \dfrac{r}{100} \right ) ^{2} -1  } = \dfrac{x^2 - 1}{x - 1}

∴ 12,066.60 × (x - 1) = 5,460 × (x² - 1) = 5,460 × (x - 1) ×(x + 1)

∴ 12,066.60 × (x - 1)/(x - 1) = 5,460 × (x + 1)

12,066.60/5,460 = x + 1

x =  12,066.60/5,460 - 1 = 1.21 = 121/100

x = 121/100

\left ( 1 + \dfrac{r}{100} \right ) ^{2} = x = \dfrac{121}{100}

1 + \dfrac{r}{100}   =\sqrt{ \dfrac{121}{100}} = \dfrac{11}{10}

We get

\dfrac{12,066.60}{5,460} =\dfrac{221}{100}

\therefore \dfrac{12,066.60}{5,460} =\dfrac{221}{100} = \left ( 1 + \dfrac{r}{100} \right ) ^{2}

1 + \dfrac{r}{100} = \sqrt{ \dfrac{221}{100} } = \dfrac{\sqrt{221} }{10}

\dfrac{r}{100} = \dfrac{\sqrt{221} }{10} - 1

\dfrac{r}{100}   = \dfrac{11}{10} - 1 = \dfrac{1}{10} = 0.1

r = 100 × 0.1 = 10%

r = 10%

Therefore, we have;

5,460 = P \cdot \left ( 1 + \dfrac{r}{100} \right ) ^{2} - P = P \times \left ( 1 + 0.1\right ) ^{2} - P

5,460  = P \times \left ( 1 + 0.1\right ) ^{2} - P = P \times \left (\left ( 1 + 0.1\right ) ^{2} - 1\right) = P \times \dfrac{21}{100}

P = \dfrac{100}{21} \times 5,460 = 26,000

The principal = Rs. 26,000

The compound interest in 3 years is therefore;

CI_3 = 26,000 \times \left ( 1 + \dfrac{10}{100} \right ) ^{3} - 26,000= 8606

The difference, 'd', between the principal and the compound interest in three years, is given as follows;

d = P - CI₃

d = 26,600 - 8606 = 17994

The difference between the principal and the compound interest in three years, d = Rs 17,994.

5 0
3 years ago
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