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valentina_108 [34]
2 years ago
10

What is the mean for 18 18 19 18 24 21 21 24 23 25 46 28 21 19

Mathematics
2 answers:
Alik [6]2 years ago
7 0
The mean is 23 (approximately).
KengaRu [80]2 years ago
5 0
(18+ 18 +19+ 18 +24+ 21+ 21 +24 +23 +25+ 46 +28+ 21 +19)/14 =325/14≈<span>23.21</span>
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Evaluate 3/2y-3+5\3 z when y =6 and z =3.
agasfer [191]

Answer:

49/4 or 12.25

Step-by-step explanation:

+ We replace y= 6 and z= 3 into this expression: 3/2y-3+5\3 z

+ We find that: 3/2y-3+5\3 z= \frac{3}{2*6} -3+5*3= \frac{1}{4} -3+15 = \frac{1}{4} +12=\frac{49}{4} = 12.25

The answer is 49/4 or 12.25

Have a good day.

6 0
3 years ago
There are 36 students in the chess club, 40 students in the drama club, 24 students in the film club. Which ratio is equivalent
zlopas [31]

Answer:

Step-by-step explanation: 40 and 24 are 5:3. We know his because:

- 40/8=5

- 24/8=3

That's how we know 40:24 = 5:3.

6 0
3 years ago
Read 2 more answers
Can someone please answer. There is one problem. There's a picture. Thank you!
swat32

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2 years ago
An athlete eats
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It is 60,000 milligrams
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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
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