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nekit [7.7K]
3 years ago
8

Statistics

Mathematics
1 answer:
Alenkasestr [34]3 years ago
4 0

Answer:

1.  mean

2. population

3. statistics

4. survey

5. mode

6. standard deviation

7. random

8. interquartile range

9. median

10. census

Step-by-step explanation:

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Find the distance between the points given.<br><br> (-1, -1) and (1, 3)<br><br> √5<br> √(17)<br> 2√5
Pani-rosa [81]

Answer:

2√5

Step-by-step explanation:

d = √(x2 - x1)² + (y2 - y1)²

= √[1 - (-1)]² + [3 - (-1)]

= √(2)² + (4)²

= √(4) + (16)

= √20

= 2√5

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Can someone find the points/coordinates of this graph?
BigorU [14]

Answer:

The x represents the value of the point on the x axis or the horizontal line, and the y represents the vertical line. Now, lets solve for the first point. We can first see that it only moves to the left by two from zero, which is basically -2. So, right now we have (-2,y). We then look for the y in which we see that it is down -6 from zero, so it will be (-2,-6). Time to look for the second point. We should get (2,-3). Now, with these two points, it is time to find the slope intercept form.  

Step-by-step explanation:

4 0
3 years ago
What is the following product?<br> 3V24•3V45 3V69 4(3V6) 6(3V5) 6(3V10)
lora16 [44]

Answer:

option 3

Step-by-step explanation:

6(3√5)

3 0
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Caleb got $10 of allowance money and babysat his cousin for $8 per hour this week. He earned a total of $46 this week. Which equ
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The answer to this question is 126 hours
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Find the exact value of tan (arcsin (two fifths)). For full credit, explain your reasoning.
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\bf sin^{-1}\left( \frac{2}{5} \right)=\theta \impliedby \textit{this simply means that}&#10;\\\\\\&#10;sin(\theta )=\cfrac{2}{5}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}\qquad \textit{now let's find the adjacent side}&#10;\\\\\\&#10;\textit{using the pythagorean theorem}\\\\&#10;c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}

\bf \pm \sqrt{5^2-2^2}=a\implies \pm\sqrt{21}=a&#10;\\\\\\&#10;\textit{we don't know if it's +/-, so we'll assume is the + one}\quad \sqrt{21}=a\\\\&#10;-------------------------------\\\\&#10;tan(\theta)=\cfrac{opposite}{adjacent}\qquad \qquad tan(\theta)=\cfrac{2}{\sqrt{21}}&#10;\\\\\\&#10;\textit{and now, let's rationalize the denominator}&#10;\\\\\\&#10;\cfrac{2}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies \cfrac{2\sqrt{21}}{(\sqrt{21})^2}\implies \cfrac{2\sqrt{21}}{21}&#10;
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