90 points to brainliest !
daniel is able to land his back flip 60% of the time. he tells his friend that in his next 5 back flips, he will land more than 3 of them.

to estimate the probability of this happening, daniel uses a computer to randomly select 5 numbers from 1 through 5. he lets 1 through 3 represent landing his back flip and 4 and 5 represent not landing it.

he repeats this for a total of 22 trials and records the results in the table below.

43113 43323 21245 53154 11545 44511 45221 14221 11123 55131 24535

35133 21454 41314 55242 52214 53542 15231 31244 54541 31415 52522
what is the best estimate, based on this simulation, of the probability that in his next 5 back flips daniel will land more than 3 of them?

27%

39%

41%

68%

The probability that in his next 5 back flips Daniel will land more than 3 of them is:

27%

Step-by-step explanation:

The outcomes that represent that he landed in his back flip more than 3 times are:

43113433231422111123       35133       15231  

So, out of the total 22 trials the number of trials he will land more than 3 of them is:

6.

Let P denotes the probability of an event.

Hence,

P(He will land more than 3 times)=Number of trials in which he landed more than three times/Total number of trials.

P( He will land more than 3 times)=6/22

Hence, the percent probability is calculated as:

Hence, the probability that in his next 5 back flips Daniel will land more than 3 of them is:

27%

answer: expected value of x = 1.0

step-by-step explanation:

p(x=0) = 0.25

p(x=1) = 0.50

p(x=2) = 0.25

expected value for x = 0×0.25 + 1×0.5 + 2×0.25 = 1.0

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length