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storchak [24]
3 years ago
8

PLEASE HELP ME WITH THIS THX!!!!!I REALLY NEED HELP WITH THIS!

Mathematics
2 answers:
Daniel [21]3 years ago
7 0
1. first point is x int., when y=0
x-4*0=-18, x=-18
(-18,0)
2. second point, y int, when x=0
0-4y=-18
-4y=-18
y=-18/-4=4.5
(0,4.5)
3. you can choose y=1, then x-4=-18, x=-14 , (-14,1)
or for this graph is
maybe better y=4, x-4*4=-18, x=-2  (-2,4)
x-4y=-18, y=3, x=-6 (-6,3)
UkoKoshka [18]3 years ago
4 0
I just want to reccomend desmos, it really helped me with graphing because im not so good at it

But what I got is: 
x-4y=-18
x=-18+4y

Root(-18,0)
Y-intercept: (0, 9/2)
You might be interested in
Consider randomly selecting a student at a certain university, and let A denote the event that the selected individual has a Vis
Mice21 [21]

Answer:

Step-by-step explanation:

Given that,

Visa card is represented by P(A)

MasterCard is represented by P(B)

P(A)= 0.6

P(A')=0.4

P(B)=0.5

P(B')=0.5

P(A∩B)=0.35

1. P(A U B) =?

P(A U B)= P(A)+P(B)-P(A ∩ B)

P(A U B)=0.6+0.5-0.35

P(A U B)= 0.75

The probability of student that has least one of the cards is 0.75

2. Probability of the neither of the student have the card is given as

P(A U B)'=1-P(A U B)

P(A U B)= 1-0.75

P(A U B)= 0.25

3. Probability of Visa card only,

P(A)= 0.6

P(A) only means students who has visa card but not MasterCard.

P(A) only= P(A) - P(A ∩ B)

P(A) only=0.6-0.35

P(A) only=0.25.

4. Compute the following

a. A ∩ B'

b. A ∪ B'

c. A' ∪ B'

d. A' ∩ B'

e. A' ∩ B

a. A ∩ B'

P(A∩ B') implies that the probability of A without B i.e probability of A only and it has been obtain in question 3.

P(A ∩ B')= P(A-B)=P(A)-P(A∩ B)

P(A∩ B')= 0.6-0.35

P(A∩ B')= 0.25

b. P(A ∪ B')

P(A ∪ B')= P(A)+P(B')-P(A ∩ B')

P(A ∪ B')= 0.6+0.5-0.25

P(A ∪ B')= 0.85

c. P(A' ∪ B')= P(A')+P(B')-P(A' ∩ B')

But using Demorgan theorem

P(A∩B)'=P(A' ∪ B')

P(A∩B)'=1-P(A∩B)

P(A∩B)'=1-0.35

P(A∩B)'=0.65

Then, P(A∩B)'=P(A' ∪ B')= 0.65

d. P( A' ∩ B' )

Using demorgan theorem

P(A U B)'= P(A' ∩ B')

P(A U B)'= 1-P(A U B)

P(A' ∩ B')= 1-0.75

P(A' ∩ B')= 0.25

P(A U B)'= P(A' ∩ B')=0.25

e. P(A' ∩ B)= P(B ∩ A') commutative law

Then, P(B ∩ A') = P(B) only

P(B ∩ A') = P(B) -P(A ∩ B)

P(B ∩ A') =0.5 -0.35

P(B ∩ A') =0.15

P(A' ∩ B)= P(B ∩ A') =0.15

5 0
2 years ago
Hector found twice as many seashells as
MissTica
Hector = 2s or h/2
Sister = s


Equation: h/2 = s

Substitution: (20)/2 = s

10 = s


Hector’s sister found 10 seashells.
4 0
3 years ago
M × m × r = 5 × 4 × 5; then r =​
Shtirlitz [24]

Answer: 5

Step-by-step explanation:

6 0
2 years ago
Find the HCF and LCM of 160 and 252 by prime factorisation method and verify that HCF x LCM = Product of the two given numbers.
MariettaO [177]

Answer:

HCF=4

LCM=10080

HCF x LCM = Product of the two given numbers is verified because 403200=40320

Step-by-step explanation:

HCF of 160 and 252

HCF is the product of common prime factors

Prime Factorisation of 160: 2x2x2x2x2x5

Prime Factorisation of 252: 2x23x3x7

HCF= 2x2

HCF=4

40 and 63 doesn't come in any other table together so,

HCF= 2 x 2 = 4

Now, finding LCM of 160 and 252

LCM is the product of all factors of both numbers, while if any factor is common, it is considered once.

Prime Factorisation of 160: 2x2x2x2x2x5

Prime Factorisation of 252: 2x23x3x7

In the given factors 2x2 is common so, it is considered once.

LCM= 2x2x2x2x2x3x5x7

LCM= 10080

Now, checking:

HCF x LCM = Product of the two given numbers.

4x10080=160x252

403200=40320

So, It is verified that HCF x LCM = Product of the two given numbers.

7 0
2 years ago
Oder the following expressions
Afina-wow [57]
A +(- c) :

a is -1 and c is 3, so it would be:
-1+(-3) = -4



c-(-b) :

c is 3 and b is 2, so it would be:
3-(-2) = 5



b :
b is 2 and there’s no addition or subtraction that needs to be done here


we have -4, 5, and 2 and going from least to greatest it’s : -4 < 2 < 5

so just look at which values gave us those answers and that would be:
a+(-c) < b < c-(-b)

so the final answer is a+(-c) < b < c-(-b)
6 0
2 years ago
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