A rectangle has a height of 7 and a width of a^4+5a^2+4a .
2 answers:
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Answer:
(a) E(Y) = 4400
sd (Y) =225
(b) P(Y ≤ 4500) = 0.67003
(c) P (X₁ > X₂) = 0.31744
Step-by-step explanation:
(a) Here we have
Y = 2·X₁ + 3·X₂
Therefore E(Y) = 2·E(X₁) + 3·E(X₂) = 2000 + 2400 = 4400
sd(Y) is given by
Variance Y = (sd (Y))² = 2²·(sd (X₁))² + 3²·(sd (X₂))²
= 4·8100 + 9·2025 = 50625
sd (Y) = √50625 = 225
(b) The probability that the revenue does not exceed 4500 is given by
P(Y ≤ 4500) = P(z ≤0.444)
z =
z = = 0.444
Therefore from the normal distribution table, we have
P = 0.67003
(c) The probability that the P(X₁ > X₂)
Since the gas station sells 2 portions of X₁ to 3 portions of X₂
Therefore, the probability that the gas station sells more of X₁ is given by
₅C₀ × 2/5⁰×3/5⁵ = 0.07776
₅C₁ × 2/5¹×3/5⁴= 0.2592
₅C₂ × 2/5²×3/5³ = 0.3456
P (X₁ > X₂) = 1 - 0.68256 = 0.31744
The group paid $ 5250 at first city and $ 6250 at second city
<u>Solution:</u>
Let x = the charge in 1st city before taxes
Let y = the charge in 2nd city before taxes
The hotel charge before tax in the second city was $1000 higher than in the first
Then the charge at the second hotel before tax will be x + 1000
y = x + 1000 ----- eqn 1
The tax in the first city was 8.5% and the tax in the second city was 5.5%
The total hotel tax paid for the two cities was $790
<em><u>Therefore, a equation is framed as:</u></em>
8.5 % of x + 5.5 % of y = 790
0.085x + 0.055y = 790 ------- eqn 2
<em><u>Let us solve eqn 1 and eqn 2</u></em>
<em><u>Substitute eqn 1 in eqn 2</u></em>
0.085x + 0.055(x + 1000) = 790
0.085x + 0.055x + 55 = 790
0.14x = 790 - 55
0.14x = 735
<h3>x = 5250</h3><em><u>Substitute x = 5250 in eqn 1</u></em>
y = 5250 + 1000
<h3>y = 6250</h3>Thus the group paid $ 5250 at first city and $ 6250 at second city