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KiRa [710]
3 years ago
9

An athlete runs every day for one week. The following week she runs one mile further than she ran on that day the previous week.

How will the mean and interquartile range for the number of miles she runs in the second week be different from those in the first week?
Mathematics
1 answer:
cluponka [151]3 years ago
5 0

Answer:

Buy $HBAR

Step-by-step explanation:

You can find it on exchanges like Bittrex, Voyager, Uphold, and soon enough Coinbase!

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While watching a football game, Jordan decided to list yardage gained as positive integers and yardage lost as negative integers
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Answer:

There is a loss of 2 yards.

Step-by-step explanation:

To find the overall gain or loss, we add or subtract there values.

Initially, we have 0 yards.

Records 14, so 0 + 14 = 14, that is, for now a gain of 14

Then loses 7, so 14 - 7 = 7, for now a gain of 7.

Then, 7 - 9 = - 2, so in all, there is a loss of 2 yards.

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C and E

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Convert 25% to a decimal
Alisiya [41]
The decimal is 0.25.
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4 years ago
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<img src="https://tex.z-dn.net/?f=%28x%5E%7B2%7D%20%2Bx-3%29%3A%20%28x%5E%7B2%7D%20-4%29%5Cgeq%201" id="TexFormula1" title="(x^{
Jlenok [28]

Answer:

x>2

Step-by-step explanation:

When given the following inequality;

(x^2+x-3):(x^2-4)\geq1

Rewrite in a fractional form so that it is easier to work with. Remember, a ratio is another way of expressing a fraction where the first term is the numerator (value over the fraction) and the second is the denominator(value under the fraction);

\frac{x^2+x-3}{x^2-4}\geq1

Now bring all of the terms to one side so that the other side is just a zero, use the idea of inverse operations to achieve this:

\frac{x^2+x-3}{x^2-4}-1\geq0

Convert the (1) to have the like denominator as the other term on the left side. Keep in mind, any term over itself is equal to (1);

\frac{x^2+x-3}{x^2-4}-\frac{x^2-4}{x^2-4}\geq0

Perform the operation on the other side distribute the negative sign and combine like terms;

\frac{(x^2+x-3)-(x^2-4)}{x^2-4}\geq0\\\\\frac{x^2+x-3-x^2+4}{x^2-4}\geq0\\\\\frac{x+1}{x^2-4}\geq0

Factor the equation so that one can find the intervales where the inequality is true;

\frac{x+1}{(x-2)(x+2)}\geq0

Solve to find the intervales when the equation is true. These intervales are the spaces between the zeros. The zeros of the inequality can be found using the zero product property (which states that any number times zero equals zero), these zeros are as follows;

-1, 2, -2

Therefore the intervales are the following, remember, the denominator cannot be zero, therefore some zeros are not included in the domain

x\leq-2\\-2

Substitute a value in these intervales to find out if the inequality is positive or negative, if it is positive then the interval is a solution, if it is negative then it is not a solution. This is because the inequality is greater than or equal to zero;

x\leq-2   -> negative

-2   -> neagtive

-1\leq x   -> neagtive

x>2   -> positive

Therefore, the solution to the inequality is the following;

x>2

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An exponent is a number written to the upper right of another number.
Romashka-Z-Leto [24]
The answer to this is question is gonna be True!
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