Answer:
100
Step-by-step explanation:
We have the sum of first n terms of an AP,
Sn = n/2 [2a+(n−1)d]
Given,
36= 6/2 [2a+(6−1)d]
12=2a+5d ---------(1)
256= 16/2 [2a+(16−1)d]
32=2a+15d ---------(2)
Subtracting, (1) from (2)
32−12=2a+15d−(2a+5d)
20=10d ⟹d=2
Substituting for d in (1),
12=2a+5(2)=2(a+5)
6=a+5 ⟹a=1
∴ The sum of first 10 terms of an AP,
S10 = 10/2 [2(1)+(10−1)2]
S10 =5[2+18]
S10 =100
This is the sum of the first 10 terms.
Hope it will help.
I am pretty sure it is 29 because i did the math and that is what it comes out to
2/3 think of this as a pie right divided into 3 area but only 2 people eat from it 1 left but how would i complete it if its not done well i need to complete the "pie"
by adding an exponent into the mesure and cacultating a mass of a=mx+b
now you divide and get 0.75
so you will fill in .25 more to get the full measurement by eating 1 more pie:
Using the Punnette Square :
R r
<span>R RR Rr </span>
<span>r Rr rr
</span>Hence,
<span>By applying the rules of probability Rr x Rr, the probability of the offspring being homozygous recessive will be one fourth or a quarter.
SO the best is :
B. 1/4</span>
Answer:
a)120
b)6.67%
Step-by-step explanation:
Given:
No. of digits given= 6
Digits given= 1,2,3,5,8,9
Number to be formed should be 3-digits, as we have to choose 3 digits from given 6-digits so the no. of combinations will be
6P3= 6!/3!
= 6*5*4*3*2*1/3*2*1
=6*5*4
=120
Now finding the probability that both the first digit and the last digit of the three-digit number are even numbers:
As the first and last digits can only be even
then the form of number can be
a)2n8 or
b)8n2
where n can be 1,3,5 or 9
4*2=8
so there can be 8 three-digit numbers with both the first digit and the last digit even numbers
And probability = 8/120
= 0.0667
=6.67%
The probability that both the first digit and the last digit of the three-digit number are even numbers is 6.67% !
