d = 3 , a₁₂ = 40 and S
= 7775
In an arithmetic sequence the nth term and sum to n terms are
<h3>• a
= a₁ + (n-1)d</h3><h3>• S
=
[2a + (n-1)d]</h3><h3>
where d is the common difference</h3><h3>a₆ = a₁ + 5d = 22 ⇒ 7 + 5d = 22 ⇒ 5d = 15 ⇔ d = 3</h3><h3>a₁₂ = 7 + 11d = 7 +( 11× 3) = 7 + 33 = 40</h3><h3>S₁₀₀ =
[(2×7) +(99×3)</h3><h3> = 25(14 + 297) = 25(311)= 7775</h3>
Answer:
6b
Step-by-step explanation:
Given:
6 girls
4 boys
Probability of 1 girl being chosen: 1/6
Probability of 1 boy being chosen: 1/4
Probability of another boy being chosen: 1/3
1/6 = 0.17 x 100% = 17%
1/4 = 0.25 x 100% = 25%
1/3 = 0.33 x 100% = 33%
17% + 25% + 33% = 75%
The probability that exactly one girl and two boys will receive awards is 75%.
The whole numbers add up to 100. So now we are left with 1/2 or 0.5, 3/5 or .60, and 1/8 or .125. Those add up to 1.225 so the amount of sand in pounds in the wheelbarrow is 101.225. Hope I helped! If you have any questions just leave a comment.
