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Alexxx [7]
2 years ago
8

The lifetimes of a certain type of automobile tire have been found to be distributed normally with a mean lifetime of 130,000 km

and a standard deviation of 1,700 km.
A) Find the Z-Score of a tire that has a lifetime of 126,005 km: Z-
(Round to two decimal places)
B) Find the lifetime of a tire if it has a z score of - 1.96: Lifetime - km
(Round to nearest whole number)
Mathematics
1 answer:
Lesechka [4]2 years ago
6 0

Answer:

b

Step-by-step explanation:

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11/12 / 2 5/8dividing fractions!!
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(11/12) / (2 5/8) = ...turn mixed numbers to improper fractions
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11/12 * 8/21 = 88/252 reduces to 22/63

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3 years ago
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Arturiano [62]

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Question 2b only! Evaluate using the definition of the definite integral(that means using the limit of a Riemann sum
lara [203]

Answer:

Hello,

Step-by-step explanation:

We divide the interval [a b] in n equal parts.

\Delta x=\dfrac{b-a}{n} \\\\x_i=a+\Delta x *i \ for\ i=1\ to\ n\\\\y_i=x_i^2=(a+\Delta x *i)^2=a^2+(\Delta x *i)^2+2*a*\Delta x *i\\\\\\Area\ of\ i^{th} \ rectangle=R(x_i)=\Delta x * y_i\\

\displaystyle \sum_{i=1}^{n} R(x_i)=\dfrac{b-a}{n}*\sum_{i=1}^{n}\  (a^2 +(\dfrac{b-a}{n})^2*i^2+2*a*\dfrac{b-a}{n}*i)\\

=(b-a)^2*a^2+(\dfrac{b-a}{n})^3*\dfrac{n(n+1)(2n+1)}{6} +2*a*(\dfrac{b-a}{n})^2*\dfrac{n (n+1)} {2} \\\\\displaystyle \int\limits^a_b {x^2} \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} R(x_i)\\\\=(b-a)*a^2+\dfrac{(b-a)^3 }{3} +a(b-a)^2\\\\=a^2b-a^3+\dfrac{1}{3} (b^3-3ab^2+3a^2b-a^3)+a^3+ab^2-2a^2b\\\\=\dfrac{b^3}{3}-ab^2+ab^2+a^2b+a^2b-2a^2b-\dfrac{a^3}{3}  \\\\\\\boxed{\int\limits^a_b {x^2} \, dx =\dfrac{b^3}{3} -\dfrac{a^3}{3}}\\

4 0
3 years ago
Find the domain for the expression:<br> y/y-5 + 1/y+8
Jet001 [13]

Answer:

1/y+4

Step-by-step explanation:

1-5+1/y+8

1/y+(1-5+8)

1/y+4

7 0
3 years ago
Read 2 more answers
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