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artcher [175]
2 years ago
10

Jake and Danny agree to share the cost of renting a house for $1100. If Jake agrees to pay $250 more than Danny, how much money

will Jake have to pay?
Mathematics
1 answer:
SVEN [57.7K]2 years ago
7 0

Step-by-step explanation:

First divide 1100 by 2

550

Now that would be the amount the both of them would have to pay if they were splitting it equally. Since there not add 250 to 550.

250+550=800

Finally subtract 250 from 550

550-250=300

Jake has to pay $550

Danny has to pay $300

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Standard for is x × 10 ^I
where,
I = indice
10 > x < 0
you cannot write this is standard form with putting in what they stand for. However you can simplify it -

4(8m-7n) +6(3n-4m)
32m - 28n + 18n - 24m
8m - 10n

but hey I am from England do what we call standard form might be different
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Is the quotient positive or negitive
ArbitrLikvidat [17]
I guess positive bcuz - -=+
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Which of the fallowing is a even number 25, 28, 31, 37
irga5000 [103]
Even numbers end with 2,4,6,8, or 0. Thus, 28 is the even number.
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3 years ago
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Need help with thisss
frozen [14]

9514 1404 393

Answer:

  cos(x) = 0.6

Step-by-step explanation:

First, we need to find UQ.

  UQ = UP·cos(y°) = UP·√(1 -sin²(y°)) = (26 cm)√(1 -25/169) = (26 cm)(12/13)

  UQ = 24 cm

Then TQ is ...

  TQ = 1/3UQ = 1/3(24 cm) = 8 cm

And cos(x°) is ...

  cos(x°) = √(1 -sin²(x)) = √(1 -64/100)

  cos(x°) = 6/10 = 3/5

  cos(x°) = 3/5 = 0.6

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}
7 0
3 years ago
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