Lionardo melly true to his name,

Mathematics

1 answer:

nordsb [41]3 years ago

7 0

Answer:

I think it is D but im not sure if you said 10 days or something else

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What the answer to this?

garik1379 [7]

Answer:

Step-by-step explanation:

$9-5\div (8-3)\times 2+6\\\\\mathrm{Follow\:the\:PEMDAS\:order\:of\:operations}\\\\\mathrm{Calculate\:within\:parentheses}\:\left(8-3\right)\::\quad 5\\=9-5\div \:5\times \:2+6\\\\\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:5\div \:5\times \:2\::\quad 2\\=9-2+6\\\\\mathrm{Add\:and\:subtract\:\left(left\:to\:right\right)}\:9-2+6\:\\:\quad 13$

7 0

3 years ago

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> $P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}$