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sergey [27]
3 years ago
13

Work out the median, range and mode​

Mathematics
1 answer:
hram777 [196]3 years ago
5 0

median= 7 middle part

range=7 because 8-1 or highest minus lowest numbers

mode= you got the answer

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Help, I need help with this math question as soon as possible!!
mina [271]

Answer:

Step-by-step explanation:

4x what = 98

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3 years ago
Marco has $75 to spend on new clothes for a job interview. He buys a shirt for $18.28, pants for $42.39, and a tie for $12.97. H
dedylja [7]

Answer:

$1.36

Step-by-step explanation:

First, add together the prices of all the clothes:

18.28 + 42.39 + 12.97

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Read 2 more answers
Write y=-1/4x-6 in standard form using integers.
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Y = -1/4 x – 6
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7 0
3 years ago
the probability of drawing a green marble from a bag of green and purple marbles is 2/7. If there are 36 green marbles in the ba
timurjin [86]

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3 years ago
Can anyone tell me why I got this wrong?
MA_775_DIABLO [31]

Answer:

The answer should be A. x \in (-\infty,\, 2).

Step-by-step explanation:

Start with the second inequality, which is more straightforward than the first. Add 3 to both sides of this equation to separate x from the numbers:

x - 3 + 3 \le -1 + 3. The direction of the inequality sign shall stay the same.

x \le 2.

On the other hand, simplifying the first inequality takes a bit more work. Start by dividing both sides by (-3). However, keep in mind that dividing or multiplying both sides of an inequality with a negative number will invert the direction of the inequality sign. For example, the left-hand side of an inequality might have been greater than the right-hand side. When both sides are multiplied with a number less than zero, the new left-hand side would become smaller than the new right-hand side.

Because (-3)\! is a negative number, dividing both sides of the first inequality inequality by (-3) will turn the "greater-than" sign into a "less-than" sign.

The inequality used to be:

-3\, (x + 2) > 15.

After dividing both sides by the negative number (-3), it becomes:

\displaystyle \frac{(-3)\, (x + 2 )}{(-3)} < \frac{15}{(-3)}.

\displaystyle (x + 2) < -5.

x < -7.

The question connected the two original inequalities with "or". Therefore, the conditions will be satisfied as long as either of at least one of the two inequalities is true. Note, that x < 2 is true whenever x is true. Therefore, any real number that is smaller than 2 will meet the requirements- not just those that are smaller than (-7).

The corresponding interval will be (-\infty,\, 2).

7 0
3 years ago
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