Let x = 78.3
y = 87.2
σₓ = 5.6 and σ₍y₎ = 6.3
number of pieces n = 50
z = z-score value =1.96 in this case
<span> A 100(1-α)%
confidence interval for the difference in large sample means is:
(x – y) ± z x </span>√( σₓ² / n+ σ₍y₎²<span>/n)
by putting the values,
=(78.3-87.2) </span>± 1.96 x √(5.6² /50 + 6.3² /50)<span>
</span><span>= -8.9 ± 2.34
</span><span>= [-11.24, -6.56]</span>
Marcos has 6 nickels and 9 quarters
<em><u>Solution:</u></em>
Let "n" be the number of nickels
Let "q" be the number of quarters
<em><u>Marcos had 15 coins in nickels and quarters</u></em>
Therefore, we can say,
number of nickels + number of quarters = 15
n + q = 15 -------- eqn 1
<em><u>He has three more quarters than nickels</u></em>
Number of quarters = 3 + number of nickels
q = 3 + n -------- eqn 2
Eqn 1 and eqn 2 represents the system of equations
<em><u>Substitute eqn 2 in eqn 1</u></em>
n + 3 + n = 15
2n + 3 = 15
2n = 15 - 3
2n = 12
<h3>n = 6</h3>
<em><u>Substitute n = 6 in eqn 2</u></em>
q = 3 + 6
<h3>q = 9</h3>
Thus Marcos has 6 nickels and 9 quarters
Answer:
Line A is the right answer.