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stich3 [128]
2 years ago
6

Anyone wanna give me stuff in rocket league pc hehehehehe

Mathematics
2 answers:
adell [148]2 years ago
7 0

Answer:

not really hehehehehe

Step-by-step explanation:

Juliette [100K]2 years ago
3 0
Maybe. What’s your name?
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Can 18/33 be simplified
oksian1 [2.3K]
Yes  because both 18 and 33 are mulyiples of 3

18/33  = 6 / 11
8 0
3 years ago
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Guys im giving max points for this, please give me a correct answer and best gets brainliest.
lukranit [14]

Answer:

y=-2/3 x +4

Step-by-step explanation:

slop = (6-0)/(-3-6)= -2/3

c= 4  (when x =0)

line equation:

y= mx + c

y=-2/3 x +4

4 0
3 years ago
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Mr Ramirez receives blueprints
densk [106]
Uhhh
Could you finish the question?
6 0
2 years ago
Jeffery babysits for 4 dollars per hour, he also works as a math tutor for 7 dollars per hour. He is only allowed to work 13 hou
Oksana_A [137]

Given:

Jeffery babysits for 4 dollars per hour and works as a math tutor for 7 dollars per hour.

He is only allowed to work 13 hours total per week.

He wants to make at least 65 dollars.

To find:

The system of inequalities to represent this situation.

Solution:

Let x be the number of hours he babysits per week and y be the number of hours h works as a math tutor.

He is only allowed to work 13 hours total per week.  It means sum of x and y must be less than or equal to 13.

x+y\leq 13

He wants to make at least 65 dollars. It means total earnings must be greater than or equal to 65.

4x+7y\geq 65

Number of hours can not be negative.

x\geq 0, y\geq 0.

Therefore, the required system of inequalities contains some equations,  x+y\leq 13,4x+7y\geq 65 and x, y\geq 0.

4 0
2 years ago
calculate the means and the standard deviation of the following set of data 4.578g, 4.581g, 4.572g, 4.573g, 4.601g, 4.577g. stat
AfilCa [17]

Answer:

68% Confidence interval  = [4.5752, 4.5848]

95% Confidence interval  = [4.5688, 4.5918]

Step-by-step explanation:

Sample mean (X) = 4.580

Sample Standard Deviation (S) = 0.01065

Sample size (n) = 6

T_{(5)} for alpha/2 0.84 = 1.1037

T_{(5)} for alpha/2 0.975 = 2.5706

68% Confidence interval = [x-T_{(5)}\frac{S}{\sqrt{n}}, x+T_{(5)]\frac{S}{\sqrt{n}}] = [4.5752, 4.5848]

95% Confidence interval = [x-T_{(5)}\frac{S}{\sqrt{n}}, x+T_{(5)}\frac{S}{\sqrt{n}}] = [4.5688, 4.5918]

7 0
3 years ago
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