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zmey [24]
3 years ago
6

Equivalent ratios

Mathematics
2 answers:
ahrayia [7]3 years ago
6 0
57:49 hdheyrbdthryebeyehegyebe
m_a_m_a [10]3 years ago
3 0
The answers are:
16:10
48:10
32:20
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22222222222222200000000
Lynna [10]

For those unaware of what this means

This is part of the thousands of reviews for a seafood place in China. Don't believe me? Look it up!

5 0
3 years ago
The following data were collected from 12 rain gauges in a park. Build a 95% CI for the mean rainfall at the park.
dybincka [34]

Answer:

Critical values:t_{\alpha/2}=-2.201 t_{1-\alpha/2}=2.201

95% confidence interval would be given by (3.646;4.472)

Step-by-step explanation:

1) Previous concepts

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 data is:

4.65 3.89 2.73 4.35 3.80 4.86 4.33 4.37 4.76 4.05 3.05 3.87

2) Compute the sample mean and sample standard deviation.

In order to calculate the mean and the sample deviation we need to have on mind the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}

=AVERAGE(4.65,3.89,2.73, 4.35, 3.8, 4.86, 4.33, 4.37, 4.76, 4.05, 3.05, 3.87)

On this case the average is \bar X= 4.059

=STDEV.S(4.65,3.89,2.73, 4.35, 3.8, 4.86, 4.33, 4.37, 4.76, 4.05, 3.05, 3.87)

The sample standard deviation obtained was s=0.6503

3) Find the critical value t* Use the formula for a CI to find upper and lower endpoints

In order to find the critical value we need to take in count that our sample size n =12 <30 and on this case we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. The degrees of freedom are given by:

df=n-1=12-1=11

We can find the critical values in excel using the following formulas:

"=T.INV(0.025,11)" for t_{\alpha/2}=-2.201

"=T.INV(1-0.025,11)" for t_{1-\alpha/2}=2.201

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

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}  

And we can use Excel to calculate the limits for the interval

Lower interval : "=4.059 -2.201*(0.6503/SQRT(12))" =3.646

Upper interval :  "=4.059 +2.201*(0.6503/SQRT(12))" =4.472

So the 95% confidence interval would be given by (3.646;4.472)

8 0
3 years ago
Sandy is helping her uncle at this coffee shop. She observes that, when ordering coffee, 13 of customers select chocolate flavor
Sophie [7]
19/50 people will probably order chocolate
3 0
3 years ago
6 and<br> Find the value of this expression if x<br> y--1.<br> xy2<br> -5
zysi [14]

Answer:

y = 5;  x = 1/5

Step-by-step explanation:

xy = 1    ----->  x = 1/y

xy^2 = 5 -----> 1/y * y^2 = 5

y^2 / y = 5

y = 5

5x = 1

x = 1/5

Hope this Helps!

5 0
3 years ago
Factoring each polynomial 4x^3+ 8x^2+ 12x
babymother [125]

Answer:4x(x^2+2x+3)

Step-by-step explanation:

6 0
3 years ago
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