A fair six-sided die is tossed 10 times, with results written in a string. The fi rst trial gave a string with only 5s and 6s. What is the probability of the 10 tosses giving a string that has only numbers greater than 4?

1/3 or 0.333333

Step-by-step explanation:

The question talks about a die that was tossed once and we are now asked to find the probability that it is a figure greater than 4

---A die has 6 faces with the numbers 1,2,3,4,5 and 6 one on each of

the six faces. Considering an unbiased die the probability of any of the faces to appear in a single roll is 1/6.

So, P(1)=P(2)=P(3)=P(4)=P(5)=P(6)=1/6

Now that numbers greater than 4 are 5 and 6

So it can be P(5) or P(6)= P(5)+P(6)=2/6 = 1/3 or 0.33333

Answer 1/3 or 0.33333

Hope this helps :)

-Danny

answer: 75°

step-by-step explanation:

∠esk is the supplement of ∠edk, so is 100°. ∠esd = 1/2·arc ed, so is 25°.

since ∠esk = ∠esd + ∠dsk

you have 100° = 25° + ∠dsk . .

fill in known values 75° = ∠dsk . . . . .

subtract 25°

step-by-step explanation: