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4 years ago

Without solving them, say whether these equations have a positive solution, a negative solution,

Mathematics

1 answer:

Mila [183]4 years ago

3 0

Answer:

a. 3x=5 Positive Solution

Reasoning: You're working with only positives

b. 5z+7=3 Negative Solution

Reasoning: If you carry the 7 over the non-coefficient numbers will be negative thus making your answer negative

c. 7-5w=3 Positive Solution

Reasoning: both the coefficients and the non coefficients will be negative if solved and thus will result in the negatives becoming positive when dividing the 5w

d. 4a=9 Positive

Reasoning: The first step would be to divide the 9 by 4 and the resulting number would be positive

e. y=y+1 Negative Solution (No solution)

Reasoning: y can never equal itself with changing. Thats like saying 0 is 1. This has no solution but based on the rules this is negative solution

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3 years ago

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3 years ago

Read 2 more answers

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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3 years ago