2(10) +5(5)= 45
b=10 a=5
Plug in the values b and a.
Y = mx + b
m = (0 - -5)/(6 - 3) = 5/3
y = 5/3x + b
0 = 5/3(6) + b
b = -10
(y = 5/3x - 10)3
3y = 5x - 30
3y - 5x = -30
5x - 3y = 30
Answer:
A. The distribution of sample means of the differences will be approximately normal if there are at least 30 years of data in the sample and/or if the population of differences in winning times for all years is normal.
Step-by-step explanation:
In other to perform a valid paired test, one of the conditions required is that, data for both groups must be approximately normal. To attain normality, the population distribution for the groups must be normal or based on the central limit theorem, the sample size must be large enough, usually n > 30. Hence, once either of the two conditions are met, the paired sample will be valid.
Simplify <span /><span><span><span><span><span><span><span><span><span>4</span><span /></span><span><span>8</span><span /></span><span><span /><span /></span></span></span></span><span /></span></span><span /></span><span>48</span></span> to <span /><span><span><span><span><span><span><span><span><span>1</span><span /></span><span><span>2</span><span /></span><span><span /><span /></span></span></span></span><span /></span></span><span /></span><span>12</span></span>
<span /><span><span><span><span><span><span><span><span><span><span>s</span><span>−</span><span>1</span></span><span /></span><span><span>4</span><span /></span><span><span /><span /></span></span></span><span>=</span><span><span><span><span>1</span><span /></span><span><span>2</span><span /></span><span><span /><span /></span></span></span></span><span /></span></span><span /></span><span>s−14=12</span></span>
2<span> </span>Multiply both sides by <span /><span><span><span><span><span><span>4</span></span><span /></span></span><span /></span><span>4</span></span>
<span /><span><span><span><span><span><span>s</span><span>−</span><span>1</span><span>=</span><span><span><span><span>1</span><span /></span><span><span>2</span><span /></span><span><span /><span /></span></span></span><span>×</span><span>4</span></span><span /></span></span><span /></span><span>s−1=12×4</span></span>
<span> </span>Simplify <span /><span><span><span><span><span><span><span><span><span>1</span><span /></span><span><span>2</span><span /></span><span><span /><span /></span></span></span><span>×</span><span>4</span></span><span /></span></span><span /></span><span>12×4</span></span> to <span /><span><span><span><span><span><span><span><span><span>4</span><span /></span><span><span>2</span><span /></span><span><span /><span /></span></span></span></span><span /></span></span><span /></span><span>42</span></span>
<span /><span><span><span><span><span><span>s</span><span>−</span><span>1</span><span>=</span><span><span><span><span>4</span><span /></span><span><span>2</span><span /></span><span><span /><span /></span></span></span></span><span /></span></span><span /></span><span>s−1=42</span></span>
Simplify <span /><span><span><span><span><span><span><span><span><span>4</span><span /></span><span><span>2</span><span /></span><span><span /><span /></span></span></span></span><span /></span></span><span /></span><span>42</span></span> to <span /><span><span><span><span><span><span>2</span></span><span /></span></span><span /></span><span>2</span></span>
<span /><span><span><span><span><span><span>s</span><span>−</span><span>1</span><span>=</span><span>2</span></span><span /></span></span><span /></span><span>s−1=2</span></span>
Add <span /><span><span><span><span><span><span>1</span></span><span /></span></span><span /></span><span>1</span></span> to both sides
<span /><span><span><span><span><span><span>s</span><span>=</span><span>2</span><span>+</span><span>1</span></span><span /></span></span><span /></span><span>s=2+1</span></span>
Simplify <span /><span><span><span><span><span><span>2</span><span>+</span><span>1</span></span><span /></span></span><span /></span><span>2+1</span></span> to <span /><span><span><span><span><span><span>3</span></span><span /></span></span><span /></span><span>3</span></span>
<span /><span><span><span><span><span><span>s</span><span>=</span><span>3</span></span><span /></span></span><span /></span><span>s=3
</span></span>
The angle congruent to ∠CGD is the angle ∠AGF as they are vertical angles.
