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zalisa [80]
3 years ago
14

I don’t get the problem pls help!

Mathematics
1 answer:
Iteru [2.4K]3 years ago
6 0
I hope this helps .......

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Solve each of these equations.<br><br> 1. 2(x+5)=3x+1<br><br> 2. 3y−4=6−2y<br><br> 3. 3(n+2)=9(6−n)
Anastasy [175]

Answer:

1. x=9

2. y=2

3. n=4

hello just helping out!

6 0
3 years ago
Which is the average rate of change over the interval [0, 10]? Hint: Find rate of change for both equations
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2 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
What is the difference between 589.01 and 0.210
gulaghasi [49]
It is 588.8 as all you have to do is subtract 0.210  from 589.01 and add up the ''.''
7 0
3 years ago
Read 2 more answers
8u^8+(-21u^8)<br> Helpppp quick
ELEN [110]

\huge\textsf{Hey there!}

\mathsf{8u^8 + (-21u^8)}\\\mathsf{= 8u^2 - 21u^8}\\\large\textsf{COMBINE the LIKE TERMS}\\\mathsf{=(8u^8 - 21u^8)}\\\mathsf{= 8 - 21\rightarrow\boxed{\bf -13}}\\\mathsf{= 8u^2 - 21u^8 \rightarrow \bf -13u^8}\\\\\\\\\boxed{\boxed{\large\textsf{Answer: \bf -13u}\bf ^2}}\huge\checkmark\\\\\\\\\large\textsf{Good luck on your assignment and enjoy your day!}\\\\\\\frak{Amphitrite1040:)}

5 0
3 years ago
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