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The imagine below is the question.<br> A Function, h(x) is defined as shown

Westkost [7]

Given function

$\\ \rm\Rrightarrow h(x)=\begin{cases}\sf \dfrac{1}{4}x-4,x\leqslant 0\\ \sf\dfrac{1}{3}x-3,0< x\leqslant 3\\ \sf \dfrac{1}{2}x-2,x\geqslant 4\end{cases}$

Take first one and put 0

$\\ \rm\Rrightarrow h(0)$

$\\ \rm\Rrightarrow {0-4}$

$\\ \rm\Rrightarrow -4$

The graph must contain (0,-4)

Only graph B contains (0,-4). so

Graph B is correct

8 0

2 years ago

Sulfur compounds cause "off-odors" in wine, so winemakers want to know the odor threshold, the lowest concentration of a compoun

sladkih [1.3K]

Answer:

a) If we construct a stemplot for the data we have this:

Stem Leaf

1 | 9

2 | 2 3 9

3 | 0 0 1 3 5

4 | 2

Notation: 1|9 means 19

b) $29.4-2.26\frac{6.75}{\sqrt{10}}=24.58$

$29.4-2.26\frac{6.75}{\sqrt{10}}=34.22$

So on this case the 95% confidence interval would be given by (24.58;34.22)

Step-by-step explanation:

For this case we have the following data:

30 30 42 35 22 33 31 29 19 23

Part a

If we construct a stemplot for the data we have this:

Stem Leaf

1 | 9

2 | 2 3 9

3 | 0 0 1 3 5

4 | 2

Notation: 1|9 means 19

As we can see the distribution is a little assymetrical to the right since we have not to much values on the right tail. But we can approximate roughly the distribution symmetric and with no outliers.

Part b

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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

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

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=29.4$

The sample deviation calculated $s=6.75$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=10-1=9$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,9)".And we see that $t_{\alpha/2}=2.26$