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dimulka [17.4K]
3 years ago
14

If f(x)= 3/x+2 - x-3 complete the following statement f(19)=

Mathematics
2 answers:
Cloud [144]3 years ago
6 0
4 1/7 or-3 6/7 i hope this helps you, you need to substitute 19 in for every x you see.
Ulleksa [173]3 years ago
5 0
-3.9–3.8
This is correct!
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Answer: 8.14

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Please help! <br> What is 58% of 18?<br> Write it out, and send picture!
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10.44

Step-by-step explanation:

58% of 18 = 10.44

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hi if anyone is good with extraneous solutions pleaseeeeeee help meeee tessa solves the equation below by first squaring both si
Sergio [31]

Answer:

x = -7/5

Step-by-step explanation:

If we square both sides of the equation, we get:

\sqrt{x^2-3x-6}=x-1\\ (\sqrt{x^2-3x-6})^2=(x-1)^2\\x^2-3x-6=x^2-2x+1\\

Then, solving for x, we get:

x^2-3x-6=x^2-2x+1\\-3x-6=2x+1\\-6-1=2x+3x\\-7=5x\\\frac{-7}{5}=x

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3 years ago
Based on the values given in the table, which triangle is a right triangle?
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2 years ago
Location is known to affect the number, of a particular item, sold by HEB Pantry. Two different locations, A and B, are selected
12345 [234]

Answer:

d) [-16.03,-3.97]

-16.03 \leq \mu_A -\mu_B \leq -3.97.

Step-by-step explanation:

Notation and 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".  

n_A=18 represent the sample of A

n_B =13 represent the sample of B

\bar x_A =39 represent the mean sample  for A

\bar x_B =49 represent the mean sample for B  

s_A =8 represent the sample deviation for A

s_B =4 represent the sample deviation for B

\alpha=0.01 represent the significance level

Confidence =99% or 0.99

The confidence interval for the difference of means is given by the following formula:  

(\bar X_A -\bar X_B) \pm t_{\alpha/2}\sqrt{(\frac{s^2_A}{n_A}+\frac{s^2_B}{n_B})} (1)  

The point of estimate for \mu_A -\mu_B is just given by:  

\bar X_A -\bar X_B =39-49=-10  

The appropiate degrees of freedom are df=n_1+ n_2 -2=18+13-2=29

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,29)".And we see that t_{\alpha/2}=2.756  

The standard error is given by the following formula:  

SE=\sqrt{(\frac{s^2_A}{n_A}+\frac{s^2_B}{n_B})}  

And replacing we have:  

SE=\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=2.188  

Confidence interval  

Now we have everything in order to replace into formula (1):  

-10-2.756\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=-16.03  

-10+2.756\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=-3.97  

So on this case the 99% confidence interval for the differences of means would be given by -16.03 \leq \mu_A -\mu_B \leq -3.97.

d) [-16.03,-3.97]

7 0
4 years ago
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