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wel
2 years ago
15

Wade Boggs played professional baseball for three different teams in 18 years. The table shows his total number of hits for each

team. Table showing the Red Sox had 2,098 hits, Yankees had 702 hits, and Devil Rays had 210 hits. Suppose he made the same number of hits each of the 18 years. If t = total number of hits and a = number of hits per year, which two equations can be used to determine the number of hits he made each year? Drag these equations into the box. t = 2,098 – 702 – 210. t = 2,098 + 702 + 210. t = 2,098 ÷ (702 + 210).a = 18 × t. a = 18 ÷ t a = t ÷ 18
Mathematics
2 answers:
Vikentia [17]2 years ago
8 0

Answer:

2

Step-by-step explanation:

none

masya89 [10]2 years ago
3 0

Answer:

answer 2

Step-by-step explanation:

none

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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
Colt1911 [192]
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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