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kolbaska11 [484]

4 years ago

8

Suppose that many large samples are taken from a population and that the sample proportions are normally distributed, with a mea

n of 0.22 and a standard error of 0.029. What is the 95% confidence interval?
by Anohaptay 12 minutes ago

1 answer:

Taya2010 [7]4 years ago

8 0

Answer:

[0.16316 , 0.27684]

Step-by-step explanation:

0.22 +/- 1.96(0.029)

0.22 +/- 0.05684

[0.16316 , 0.27684]

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Scientists have found a relationship between the temperature and the height above a distant planet's surface. , given below, is

ikadub [295]

The question is incomplete, the complete question is below:

Scientists have found a relationship between the temperature and the height above a distant planet's surface. , given below, is the temperature in Celsius at a height of kilometers above the planet's surface. The relationship is as follows: T(h) = 30.5 -2.5h.

a) Calculate T^-1(x)

b) T^-1(15)

Answer:

a) $T^{-1} (x)= \frac{30.5-x}{2.5}$

b) 6.2 °C

Step-by-step explanation:

a) $T(h) = 30.5 - 2.5h\\2.5h = 30.5-T(h)\\h = \frac{30.5-T(h)}{2.5}\\ T^{-1} (x)= \frac{30.5-x}{2.5}$

b) $T^{-1} (x)= \frac{30.5-x}{2.5}\\\\T^{-1} (15)= \frac{30.5-15}{2.5}=6.2$

3 0

3 years ago

Is 0.05=1/2 correct?

BlackZzzverrR [31]

Answer:

No

Step-by-step explanation:

To convert a decimal to a fraction you multiply by one hundred. Then, you write that number as a fraction over 100 and simplify. Eg 0.5×200 = 50 , 50/100 simplifies to 1/2. Therefore 0.05 as a fraction is 0.05×100 = 5, 5/100 simplified = 1/20

4 0

3 years ago

What is the y-intercept of the quadratic function f(x) = (x – 8)(x + 3)?

frez [133]

Answer:

y-intercept is -24

Step-by-step explanation:

The solution is in the picture

8 0

4 years ago

PLEASE ANSWER MY QUESTEN THAT IS SHOWN BELOW

Reika [66]

Answer:

$2^{3} *3^{3}$

Step-by-step explanation:

8 0

3 years ago

The mean weight of 500 college students is 70 kg and the standard deviation is 3 kg. Assuming that the weight is normally distri

Anna71 [15]

Answer: 4.75%.

Step-by-step explanation:

Given: The mean weight $(\mu)$ of 500 college students is 70 kg and the standard deviation $(\sigma)$ is 3 kg.

Let X be the number of students weight over 75 kg.

If the weight is normally distributed, then the probability that students weigh over 75 kg:

$P(X>75)=P(\dfrac{X-\mu}{\sigma}>\dfrac{75-70}{3})\\\\=P(z>\dfrac53)\\\\=P(z>1.67)\\\\=1-P(z$

Hence, the percentage of students weigh over 75 kg = 4.75%.

6 0

3 years ago