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notsponge [240]
3 years ago
10

PLEASE HELP GIVING BRAINLIEST 25 POINTS!

Mathematics
2 answers:
tatiyna3 years ago
8 0

Answer:

nearest tenth= 5.4

Step-by-step explanation:

55 ÷ 10.1= 5.44554455446.

Tom [10]3 years ago
4 0

Answer:

<h2><u><em>5.4</em></u></h2>

Step-by-step explanation:

55/10.1= 5.44554455445544554

Round to the nearest tenth

4 is less so, 5.4

<h2>5.4 is the answer</h2>
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Please help will mark brainliest
Wewaii [24]
Since this is a triangle all the angles within the triangle add up to 180 degrees so....
82+(9x+2)+(7x)=180
82+2+16x=180
16x+84=180
16x=96
X=6
If we plug in X to angle(9x+2) and angle (7x) we will get 56 and 42. We can get angle 2 by subtracting 180-42 which is 138
FINAL ANSWER: 138 degrees
4 0
3 years ago
Solve the system using substitution. <br><br> 2x + 2y = 30<br> y - 4x = 0<br><br> ( _ , _ )
Kay [80]

Answer:

x = 3

y= 12

Step-by-step explanation:

2x + 2y = 30

y - 4x = 0  /*2

2x + 2y = 30

2y - 8x = 0

-----------------

10x = 30

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7 0
3 years ago
Read 2 more answers
PLEASE HELP ME WITH THIS!!!!!!
Lapatulllka [165]

Answer:

Parallelogram

Step-by-step explanation:

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6 0
3 years ago
How many times does 59 go into 6
sveticcg [70]
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4 0
4 years ago
Read 2 more answers
43 milk jugs are in a fridge with 11 that are spoiled. 5 jugs are randomly selected. what are the odds that the first two are go
malfutka [58]

Answer:

0.85% probability that the first two are good and the last three are spoiled

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the first two jugs is not important, as is not the order in which the last three are selected. So we use the combinations formula to solve this question.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

Desired outcomes:

2 spoiled, from a set of 43-11 = 32.

3 spoiled, from a set of 11.

So

D = C_{32,2}*C_{11,3} = \frac{32!}{2!(32-2)!}*\frac{11!}{3!(11-3)!} = 982080

Total outcomes:

5 jugs from a set of 43.

T = C_{43,5} = \frac{43!}{5!(43-5)!} = 115511760

Probability:

p = \frac{D}{T} = \frac{982080}{115511760} = 0.0085

0.85% probability that the first two are good and the last three are spoiled

4 0
3 years ago
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