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aleksley [76]
3 years ago
15

one method for deciding between two chioces is to toss a coin. what are all the possible outcomes when you toss a coin?

Mathematics
2 answers:
Arte-miy333 [17]3 years ago
5 0
Either heads or tails so there is a 50/50 chance of both
Elena-2011 [213]3 years ago
4 0
If you were to toss a coin, the possible outcomes would be: heads or tails.
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HURRY HURRY (function)
Semmy [17]

Answer:

h

Step-by-step explanation:

6 0
3 years ago
What is the ordered pair for y=(1/3)x-4<br>A.(3,-3)<br>b.(6,4)<br>c.(1,-4)<br>d.(8,-5)​
labwork [276]

Answer:

The answer is (3,-3)

Step-by-step explanation:

y=(1/3)x-4

simplify:

y= (.3*x) -4

substitute:

-3= (.3*3) -4

solve:

PEMDAS

-3= (.3*3) -4

-3= .9 - 4

-3= -3.1

although this is not the exact same it is the closest out of all the equations therefore the best answer you have

7 0
3 years ago
What is 1/2x16(4+v)=136
lana66690 [7]
6.5
-----------------------
5 0
3 years ago
Read 2 more answers
If you have to apply 30 n of force on a crowbar to lift an object that weighs 330 n, what is the mechanical advantage of the cro
shutvik [7]
<span>The answer is: 2. 11 Crowbar is a Class A lever Therefore the mechanical advantage of the crowbar can be given by either: MA = Output distance/Input distance OR MA = Output force/Input force Since, the question gives only the force, we can use the second formula. MA = Output force/ Input Force = 330 N / 30 N = 11</span>
5 0
3 years ago
The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard dev
Vladimir [108]

Answer:

Probability that the average length of a sheet is between 30.25 and 30.35 inches long is 0.0214 .

Step-by-step explanation:

We are given that the population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches.

Also, a sample of four metal sheets is randomly selected from a batch.

Let X bar = Average length of a sheet

The z score probability distribution for average length is given by;

                Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean = 30.05 inches

           \sigma   = standard deviation = 0.2 inches

             n = sample of sheets = 4

So, Probability that average length of a sheet is between 30.25 and 30.35 inches long is given by = P(30.25 inches < X bar < 30.35 inches)

P(30.25 inches < X bar < 30.35 inches)  = P(X bar < 30.35) - P(X bar <= 30.25)

P(X bar < 30.35) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{30.35-30.05}{\frac{0.2}{\sqrt{4} } } ) = P(Z < 3) = 0.99865

 P(X bar <= 30.25) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{30.25-30.05}{\frac{0.2}{\sqrt{4} } } ) = P(Z <= 2) = 0.97725

Therefore, P(30.25 inches < X bar < 30.35 inches)  = 0.99865 - 0.97725

                                                                                       = 0.0214

                                       

7 0
3 years ago
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