Let us compute first the probability of ending up an odd number when rolling a dice. A dice has faces with numbers 1 up to 6. The odd numbers within that is 3 (1, 3 and 5). Therefore, each dice has a probability of 3/6 or 1/2. Then, you use the repeated trials formula:
Probability = n!/r!(n-r)! * p^r * q^(n-r), where n is the number of tries (n=6), r is the number tries where you get an even number (r=0), p is the probability of having an even face and q is the probability of having an odd face.
Probability = 6!/0!(6!) * (1/2)^0 * (1/2)^6
Probability = 1/64
Therefore, the probability is 1/64 or 1.56%.
<u>Solution </u><u>1</u><u> </u><u>:</u><u>-</u>
Given expression ,
We know that,
Therefore , this can be written as ,
![\implies \sqrt[4]{13^3}](https://tex.z-dn.net/?f=%5Cimplies%20%5Csqrt%5B4%5D%7B13%5E3%7D%20)
<u>Solution</u><u> </u><u>2</u><u>:</u><u>-</u>
Given expression ,
We know that,
Therefore, this can be written as ,

Answer:
347
Step-by-step explanation:
i added them all
I am not 100% positive but from what I remember learning I think the answer you are looking for on the first question is C.
If it is circle and the circle have the center at point O then the OP=1/2of length AB.