there are 6 terms! hope this helped
For the answer to the question above, I'll provide my solutions to my answers for the problem below.
(–2x3y2 + 4x2y3 – 3xy4) – (6x4y – 5x2y3 – y5)
(−2x3)(y2)+4x2y3+−3xy4+−1(6x4y)+−1(−5x2y3)+−1(−y5)
(−2x3)(y2)+4x2y3+−3xy4+−6x4y+5x2y3+y5
−2x3y2+4x2y3+−3xy4+−6x4y+5x2y3+y5
−2x3y2+4x2y3+−3xy4+−6x4y+5x2y3+y5
(−6x4y)+(−2x3y2)+(4x2y3+5x2y3)+(−3xy4)+(y5)
−6x4y+−2x3y2+9x2y3+−3xy4+y5
So the answer is,
= <span><span><span><span><span>−<span><span>6x4</span>y</span></span>−<span><span>2x3</span>y2</span></span>+<span><span>9x2</span>y3</span></span>−<span>3xy4</span></span>+y5</span>
I hope this helps
Complete question :
Birth Month Frequency
January-March 67
April-June 56
July-September 30
October-December 37
Answer:
Yes, There is significant evidence to conclude that hockey players' birthdates are not uniformly distributed throughout the year.
Step-by-step explanation:
Observed value, O
Mean value, E
The test statistic :
χ² = (O - E)² / E
E = Σx / n = (67+56+30+37)/4 = 47.5
χ² = ((67-47.5)^2 /47.5) + ((56-47.5)^2 /47.5) + ((30-47.5)^2/47.5) + ((37-47.5)^2/47.5) = 18.295
Degree of freedom = (Number of categories - 1) = 4 - 1 = 3
Using the Pvalue from Chisquare calculator :
χ² (18.295 ; df = 3) = 0.00038
Since the obtained Pvalue is so small ;
P < α ; We reject H0 and conclude that there is significant evidence to suggest that hockey players' birthdates are not uniformly distributed throughout the year.
Answer:
4) Alternate Interior angles 5) Parallel lines property.
Step-by-step explanation:
The question is asking us to Complete the statements to prove that line AB ⩭ to line CD and line BC ⩭ to line AD.
In statement 4 .∠CAB is congruent to ∠ACD as AB is parallel to CD and ∠BCA is congruent to∠CAD as AD is parallel to BC and these are Alternate interior angles to the parallel lines .
In statement 5.m∠CAB =∠ACD and ∠BCA = ∠CAD as by property of parallel lines Alternate interior angles are equal.
