3x< 21 fill in the blank x < ____

1 answer:

7 0

To solve this problem, you need to have x alone on one side of the inequality. To do this, you need to undo the multiplication that is bonding 3 and x together. To undo this, you need to divide both sides by 3.

This results in the inequality x<7.

Your final answer is x<7.

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For "downhill" lines, if the rise is negative, the run must be<br> ?

Naya [18.7K]

9514 1404 393

Answer:

positive

Step-by-step explanation:

When considering the slope of a line, we usually choose "rise" and "run" so that "run" is positive, and "rise" measures the vertical change as one moves from left to right.

If we assume that "downhill" lines refers to lines that go from upper left to lower right (downhill to the right), then the rise is negative while the run is positive.

<em>Additional comment</em>

Here, we assume a line is considered to be "downhill" if it has negative slope. In the real world, "downhill" is the direction the ball rolls under the influence of gravity. That direction might be behind you, or to your left, instead of in front of you or to the right.

Definitions are important in math. "Downhill" is not a term that is widely used to describe lines, so we have to guess the intended meaning here.

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Simplify: (9w9 + 4w - 4)+(2w9 +6w +9)

kirill115 [55]

Answer: Its 11w9+10w+5

Step-by-step explanation:

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3 years ago

A box in a supply room contains 24 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 7 are

Marrrta [24]

Answer:

a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

b) There is a 8.65% probability that all three of the bulbs have the same rating.

c) There is a 12.45% probability that one bulb of each type is selected.

Step-by-step explanation:

There are 24 compact fluorescent lightbulbs in the box, of which:

8 are rated 13-watt

9 are rated 18-watt

7 are rated 23-watt

(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?

There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.

It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So

$P = p^{3}_{2,1}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = \frac{3!}{2!1!}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 3*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 0.1764$