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son4ous [18]
3 years ago
12

A repair bill for your car is $553. The parts cost $265. The labor cost is $48 per hour. Write and solve an equation to find the

number of hours of labor spent repairing the car.
Mathematics
1 answer:
rusak2 [61]3 years ago
8 0
(553-265)/48=6
6 hours spent repairing the car
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The expression 3x^3-5x^2+3x-1 has ___ terms and a constant of ____.
Andrews [41]
<span>The expression 3x^3-5x^2+3x-1 has ___ terms and a constant of ____.
have four terms and 1 constant
</span>
5 0
4 years ago
X ⋅ x = ? (1 point) jjjj
liraira [26]

Answer:

x-x = 0

x+x= 2x

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have a nice day...

4 0
2 years ago
An instructor who taught two sections of Math 161A, the first with 20 students and the second with 30 students, gave a midterm e
uranmaximum [27]

Answer:

<em>The answers are for option (a) 0.2070  (b)0.3798  (c) 0.3938 </em>

Step-by-step explanation:

<em>Given:</em>

<em>Here Section 1 students = 20 </em>

<em> Section 2 students = 30 </em>

<em> Here there are 15 graded exam papers. </em>

<em> (a )Here Pr(10 are from second section) = ²⁰C₅ * ³⁰C₁₀/⁵⁰C₁₅= 0.2070 </em>

<em> (b) Here if x is the number of students copies of section 2 out of 15 exam papers. </em>

<em>  here the distribution is hyper-geometric one, where N = 50, K = 30 ; n = 15 </em>

<em>Then, </em>

<em> Pr( x ≥ 10 ; 15; 30 ; 50) = 0.3798 </em>

<em> (c) Here we have to find that at least 10 are from the same section that means if x ≥ 10 (at least 10 from section B) or x ≤ 5 (at least 10 from section 1) </em>

<em> so, </em>

<em> Pr(at least 10 of these are from the same section) = Pr(x ≤ 5 or x ≥ 10 ; 15 ; 30 ; 50) = Pr(x ≤ 5 ; 15 ; 30 ; 50) + Pr(x ≥ 10 ; 15 ; 30 ; 50) = 0.0140 + 0.3798 = 0.3938 </em>

<em> Note : Here the given distribution is Hyper-geometric distribution </em>

<em> where f(x) = kCₓ)(N-K)C(n-x)/ NCK in that way all these above values can be calculated.</em>

7 0
3 years ago
If the instantaneous velocity of an object is zero, then its acceleration must be:
Nuetrik [128]
Hi I think it would be zero sense the object is in one spot. I hope I helped
6 0
4 years ago
Find the least common multiple 36, 27 and 10
Ostrovityanka [42]
36= 2*2*3*3
27= 3*3*3
10= 2*5

LCM= 2*2*3*3*3*5
LCM= 540

Final answer: 540
4 0
3 years ago
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