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3 years ago

When was pi discovered

2 answers:

8 0

Pi was discovered way back already the way to Babylonians and Egyptians 216 BC but none were all the way accurate they were close

Scilla [17]3 years ago

3 0

Answer:

As early as 1900 BC, geometers from ancient civilizations such as Babylon, India, and Egypt, are already aware of one fundamental property of the circle: that a circle’s circumference is always a little over three times that of its diameter. A more precise value was not available however as each of the civilizations computed pi only within 1% of its actual value.

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Expand and combine like terms. (7m6 + 6)(7m6 – 6)

Darina [25.2K]

5+620=728-72=*8%fc per second

4 0

3 years ago

The Ohio Department of Agriculture tested 203 fuel samples across the state

Rus_ich [418]

Answer:

$\hat p = \frac{14}{105}= 0.133$

And that represent the proportion of failures.

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.133 - 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.0475$

$0.133 + 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.2185$

The 99% confidence interval would be given by (0.0475;0.2185)

Step-by-step explanation:

Previous concept

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by:

$z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58$

The proportion estimated would be:

$\hat p = \frac{14}{105}= 0.133$

And that represent the proportion of failures.

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.133 - 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.0475$

$0.133 + 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.2185$

The 99% confidence interval would be given by (0.0475;0.2185)

3 0

3 years ago

Read 2 more answers

What is the percent increase if your salary went from $85 to $100 per week? Round your answer to the nearest percent, if necessa

slavikrds [6]

% increase = 100 - 85
--------- * 100
85

= 17.65
= 18 % to nearest per cent

3 0

3 years ago

Read 2 more answers

The sum of two numbers is 49 and the difference is 25 . What are the numbers?

OLEGan [10]

Answer:

x = 37 y = 12

Step-by-step explanation:

Let x be one number

Let y = other number

x+y = 49

x-y = 25

Add the equations together

x+y = 49

x-y = 25

---------------

2x = 74

Divide by 2

2x/2 = 74/2

x =37

x+y = 49

37+y = 49

Subtract 37 from each side

37+y-37 = 49-37

y = 12

4 0

3 years ago

The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds.Refer to

Dafna11 [192]

Given that

the weight of football players is distributed with a mean of 200 pounds and a standard deviation of 25 pounds.

And we need to find What is the minimum weight of the middle 95% of the players?

Explanation -

Using the Empirical Rule, 95% of the distribution will fall within 2 times of the standard deviation from the mean.

Two standard deviations = 2 x 25 pounds = 50 pounds

So the minimum weight = 200 pounds - 50 pounds = 150 pounds

Hence the final answer is 150 pounds.

3 0

1 year ago