Just divide 121 by 11 to get your answer.
(5x - 3y =23) + (7x +3y = -11) =
12x = 12
x = 1
Substitute
5(1) -3y =23
5 - 3y = 23
-3y = 18
y = -6
x=1 and y=-6
Part (a)
<h3>
Answer: Ø</h3>
This is the empty set
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Explanation:
It doesn't matter what set A is composed of. Intersecting any set with the empty set Ø will always result in the empty set.
This is because we're asking the question: "What does some set A and the empty set have in common?". The answer of course being "nothing" because there's nothing in Ø. Not even the value zero is in this set.
We can write Ø as { } which is a set of curly braces with nothing inside it.
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Part (b)
<h3>Answer: {1,2,3,4,5,6}</h3>
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Explanation:
When you union the universal set with any other set, you'll get the universal set.
The rule is 
where I've made B the universal set to avoid confusion of the letter U and the union symbol 
which looks nearly identical.
Why does this rule work? Well if an item is in set 
, then it's automatically in set U (everything is in set U; it's the universe). So we're not adding anything to the universe when applying a union involving this largest set.
It's like saying
- A = set of stuff inside a persons house
- = set of stuff outside a persons house (ie stuff that is not in set A)
- U = set of every item
we can see that 
will basically form the set of every item, aka the universal set.
Answer:
- 9/2
Step-by-step explanation:
slope m = (y2 - y1) /(x2 - x1) = (14 - 5)/(1 - 3)= 9/(-2)= - 9/2
Complete question :
It is desired to check the calibration of a scale by weighing a standard 10 g weight 100 times. Let µ be the population mean reading on the scale, so that the scale is in calibration if µ = 10. A test is made of the hypotheses H0:µ = 10 versus H1:µ ≠ 10. Consider three possible conclusions: (i) The scale is in calibration, (ii) The scale is out of calibration, (iii) The scale might be in calibration.
Which of the three conclusions is best ifH0 is rejected?
Answer:
The scale is out of calibration
Step-by-step explanation:
The null hypothesis ; H0
H0 : μ = 10 ; if this hold true, then the scale is in calibration
Alternative hypothesis ; H1
H1 : μ ≠ 10 ; if this holds true, then the scale is out of calibration
If the Null hypothesis, H0 is rejected, it means that, there is significant evidence to support the alternative hypothesis ; that the scale is out of calibration.
Hence, the best conclusion is that, the scale is out of calibration.
