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%.
according to the question
3x-1=0
3x=1
x=⅓
so
f(x)=18x³+x-1
f(⅓)=18.(⅓)³+⅓-1
f(⅓)=18.⅓.⅓.⅓+⅓-1
f(⅓)=6.⅑+⅓-1
f(⅓)=⅔+½-1
f(⅓)=0
<h3>therefore</h3><h3> the remainder is 0</h3>
Roots stems finite solution
Answer:
r = 7 q - 34
Step-by-step explanation:
Solve for r:
-34 + 7 q - r = 0
Subtract 7 q - 34 from both sides:
-r = 34 - 7 q
Multiply both sides by -1:
Answer: r = 7 q - 34
15/8 + 1/4
15/8 + 2/8
= (15 + 2) /8
= 17/8
= 2 1/8. 2 whole number 1/8.
