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natta225 [31]
3 years ago
5

A group of children, 6 to 10 years old, were asked how many video games they owned. The scatter plot shows the results.

Mathematics
1 answer:
eimsori [14]3 years ago
3 0
The answer is 10.<span>Think of the outlier as in the dot that's off-topic out of the entire graph. </span>
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A toy manufacturer wants to know how many new toys children buy each year. A sample of 305 children was taken to study their pur
Harrizon [31]

Answer:

The 80% confidence interval for the mean number of toys purchased each year is between 7.5 and 7.7 toys.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.8}{2} = 0.1

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.1 = 0.9, so Z = 1.28.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.28\frac{1.5}{\sqrt{305}} = 0.1

The lower end of the interval is the sample mean subtracted by M. So it is 7.6 - 0.1 = 7.5

The upper end of the interval is the sample mean added to M. So it is 7.6 + 0.1 = 7.7

The 80% confidence interval for the mean number of toys purchased each year is between 7.5 and 7.7 toys.

8 0
3 years ago
I need help simplifying -14 1/5-(-13 1/10)
xxMikexx [17]

Answer:

-141/5-(131/10)= -41.3

4 0
3 years ago
I need help with this problem , GEOMETRY
34kurt

k:\ y=m_1x+b_1\\l:\ y=m_2x+b_2\\\\k\ ||\ l\iff m_1=m_2\\\\k\ \perp\ l\iff m_1m_2=-1

1.\\(-6,\ 1),\ (0,\ -2)\\\\m_1=\dfrac{-2-1}{0-(-6)}=\dfrac{-3}{6}=-\dfrac{1}{2}\\\\(0,\ 3),\ (4,\ 1)\\\\m_2=\dfrac{1-3}{4-0}=\dfrac{-2}{4}=-\dfrac{1}{2}\\\\m_1=m_2\\\\Answer:\ Parallel


(-4,\ 0),\ (2,\ 4)\\\\m_1=\dfrac{4-0}{2-(-4)}=\dfrac{4}{6}=\dfrac{2}{3}\\\\(-3,\ 4),\ (1,\ -2)\\\\m_2=\dfrac{-2-4}{1-(-3)}=\dfrac{-6}{4}=-\dfrac{3}{2}\\\\m_1m_2=\dfrac{2}{3}\cdot\left(-\dfrac{3}{2}\right)=-1\\\\Answer:\ Perpendicular

5 0
3 years ago
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