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alex41 [277]
3 years ago
5

Solve the inequality. Check your solutions.

Mathematics
1 answer:
Komok [63]3 years ago
8 0
The inequality is 

7- \frac{2}{b}\ \textless \ \frac{5}{b}

write 7 as 7b/b to have all the expressions in common denominator:

\frac{7b}{b} - \frac{2}{b}\ \textless \ \frac{5}{b}

\frac{7b-2}{b} \ \textless \ \frac{5}{b}

\frac{7b-2}{b}- \frac{5}{b}\ \textless \ 0

\frac{7b-2-5}{b}\ \textless \ 0

\frac{7b-7}{b}\ \textless \ 0

\frac{7(b-1)}{b}\ \textless \ 0

here b=1 is a root and b=0 is not in the domain of the expression, but it still has an effect in the sign of the expression.

the sign table of \frac{7(b-1)}{b} is :

+++++++[0] --------[1] +++++

this means that for values of b to the left of 0 and to the right of 1, the expression is positive, and for values of b in (0, 1), the expression is negative.

that is \frac{7(b-1)}{b}\ \textless \ 0 for b∈(0, 1)

Answer: (0, 1)
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Average precipitation for the first 7 months of the year, the average precipitation in toledo, ohio, is 19.32 inches. if the ave
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Part A:

The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \  \frac{a-\mu}{\sigma}\right)

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:

P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\  \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\  \\ =1-P(z\ \textless \ -0.5410) \\  \\ =1-0.29426=\bold{0.7057}



Part B:

</span>The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:

</span>P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}
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3 years ago
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