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worty [1.4K]
2 years ago
10

Draw a model to show 5.5/5

Mathematics
1 answer:
REY [17]2 years ago
7 0
I don't understand the question you are asking sorry.
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The equations below represent the numbers y of tickets sold after x weeks for two different local music festivals
Anton [14]
Assuming you mean y=10x+150 and y=20x+115, you need to use a simultaneous equation, because you have two equations with two unknowns (x and y)

rearrange so
10x-y=-150
20x-y=-115

multiply the top by -1, so that if we add the two lines together, the y will cancel out
-10x+y=150
20x-y=-115

add the two lines together
10x=35
x=3.5

so the time is 3 and a half weeks

then we can sub in x to find y
20x-y=-115
20(3.5)-y=-115
70-y=-115
-y=-185
y=185

so 185 tickets were sold !
you can sub these values into your original equations to check your answer :)
5 0
2 years ago
Read 2 more answers
The distribution of average wait times in drive-through restaurant lines in one town was approximately normal with mean \mu = 18
Fynjy0 [20]

Answer:

193

Step-by-step explanation:

8 0
2 years ago
The weight y of an object on Jupiter is proportional to the weight x of the object on Earth. An object that weighs 150 pounds on
svetoff [14.1K]
Weight of an object=J
x=J/(379.2/150)
x=J/2.528

y=12.64/2.528
y=5

Hope this helps :)
7 0
2 years ago
Read 2 more answers
Ms. Lopez draws two cylinders on the whiteboard. The first cylinder has a diameter of 6 inches and a height of 14 inches. The se
chubhunter [2.5K]

Answer:

7 inches

Step-by-step explanation:

Height : Diameter

14 : 6

H : 3

H/14 = 3/6

H = 14 × 3/6

H = 7

7 0
2 years ago
Discuss the validity of the following statement. If the statement is always​ true, explain why. If​ not, give a counterexample.
Zarrin [17]

Correction:

Because F is not present in the statement, instead of working on​P(E)P(F) = P(E∩F), I worked on

P(E∩E') = P(E)P(E').

Answer:

The case is not always true.

Step-by-step explanation:

Given that the odds for E equals the odds against E', then it is correct to say that the E and E' do not intersect.

And for any two mutually exclusive events, E and E',

P(E∩E') = 0

Suppose P(E) is not equal to zero, and P(E') is not equal to zero, then

P(E)P(E') cannot be equal to zero.

So

P(E)P(E') ≠ 0

This makes P(E∩E') different from P(E)P(E')

Therefore,

P(E∩E') ≠ P(E)P(E') in this case.

8 0
2 years ago
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