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Brilliant_brown [7]
3 years ago
9

For the following​ case, assume that the population standard deviation is known and decide whether use of the​ z-interval proced

ure to obtain a confidence interval for the population mean is reasonable. Explain your answer. The variable under consideration is very close to being normally​ distributed, and the sample size is 10. Choose the correct answer below. A. It is not reasonable to use the​ z-interval procedure in this case since it is very likely that with such a small sample size that there will be outliers in the sample. Knowing that the variable under consideration is very close to being normally distributed does not mean the sample is. B. It is reasonable to use the​ z-interval procedure in this case since both the sample is large​ (size 10 or​ greater) and the variable under consideration is very close to being normally distributed. C. It is reasonable to use the​ z-interval procedure in this case​ since, although the sample is small​ (size less than​ 15), the variable under consideration is very close to being normally distributed. D. It is not reasonable to use the​ z-interval procedure in this case because the sample is too small​ (size less than​ 15).
Mathematics
1 answer:
Ray Of Light [21]3 years ago
8 0

Answer:

B. It is reasonable to use the? z-interval procedure in this case? since, although the sample is small? (size less than? 15), the variable under consideration is very close to being normally distributed.

Step-by-step explanation:

answer b is considered to be correct because we know that the population is normal and the standard deviation is known, which allows using the interval z, the answer A is not correct because although the option to use the interval z is given, which is correct, the large sample is not favored, the answer C and D are incorrect because they both reject the use of the z interval and in d it is further rejected that although there is a normal distribution the sample is not, which is false

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At 3pm, the length of the shadow of a thin vertical pole standing on level ground is the same as the height of the pole. A while
aliya0001 [1]

Answer: The height of the pole is 175.97 ( approx )

Step-by-step explanation:

Let the height of the pole is x cm,

Also, let the angle of elevation of the sun to the pole at 3 pm is \theta

Thus, by the question,

tan\theta = \frac{\text{ The height of the pole}}{\text{ The height of the shadow}}

\implies tan\theta = \frac{x}{x}    ( at 3pm the height of shadow = height of the pole)

\implies tan\theta = 1

\implies \theta = 45^{\circ}

Again, according to the question,

When the angle of elevation is   (\theta - 12^{\circ}),

The height of shadow = x + 95,

\implies tan(45-12)^{\circ}=\frac{x}{x+95}

\implies tan 33^{\circ}=\frac{x}{x+95}

\implies tan 33^{\circ}x + 95\times tan33^{\circ}=x

\implies tan 33^{\circ}x - x = - 95\times tan33^{\circ}

\implies -0.3505924068 x = - 61.6937213538

x=175.969930202\approx 175.97\text{ cm}

4 0
3 years ago
a ball is shot straight upward. with it's height, in feet, after t seconds given by the function f(t)=-16t^2+192t. Find the aver
jekas [21]

ANSWER

80  {ms}^{ - 1}

EXPLANATION

The average velocity of the ball is the rateof displacement over the total time.

The height of the ball, in feet, after t seconds is given by the function:

f(t)=-16t^2+192t

At time t=1, the height of the ball is

f(1)=-16(1)^2+192(1)

f(1)=-16+192 = 176ft

At time t=6, the height of the ball is

f(6)=-16(6)^2+192(6)

f(6)=-16(36)+192(6)

f(6)=-576+1152 = 576

The average velocity

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=  \frac{576- 176}{6 - 1}

=  \frac{400}{5}

= 80  {ms}^{ - 1}

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Answers:

a= 5x±\sqrt{25x^{2} +24u^{2} \\

 ------------------------------

                6  

u=± \sqrt{2a(3z-5x}

  -------------------------

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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