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jenyasd209 [6]
2 years ago
10

Amy graphed a point to represent the absolute value of a number on a number line. If the original number is less than −10, descr

ibe all the possible values for the point Amy graphed on the number line.
Mathematics
1 answer:
Sloan [31]2 years ago
5 0

Answer:

x>= 11 or x>10

Step-by-step explanation:

Absolute value can be used to represent the positive counterpart of a negative number. In this case, we are asked to find a number that is less than -10, meaning from -11 to negative infinity. Since we are asked to take absolute value, we would take |-11| which equals 11, and |-∞| = ∞.

So Amy graphed a value on the number line that is greater than or equal to 11, or any value greater than 10.

*Infinity is not a real number, it represents the largest number possible, which doesn't exist since numbers can continue on forever, so it can be disregarded.*

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Solve for x. Show each step of the solution.
azamat

Answer:

simplify 36 - 4.5x + 36 to -4.5x + 72

simplify 102 - 7.5x - 60 to -7.5x + 42

add 7.5x to both sides

simplify -4.5x + 72 + 7.5x to 3x + 72

subtract 72 from both sides

simplify 42 - 72 to -30

divide both sides by 3

simplify 30/3 to 10

Answer: x = -10

8 0
3 years ago
If the 8% tax on a sale amounts to 96. What is the final price (tax included) of the item?
GenaCL600 [577]

I believe it's E.) 12.96


12 times 8% equals .96 so $12.96 should be the answer since it wants the final price with tax :DD

8 0
3 years ago
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. A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs ar
yaroslaw [1]

Answer:

a) 59.34%

b) 44.82%

c) 26.37%

d) 4.19%

Step-by-step explanation:

(a)

There are in total <em>4+5+6 = 15 bulbs</em>. If we want to select 3 randomly there are  K ways of doing this, where K is the<em> combination of 15 elements taken 3 at a time </em>

K=\binom{15}{3}=\frac{15!}{3!(15-3)!}=\frac{15!}{3!12!}=\frac{15.14.13}{6}=455

As there are 9 non 75-W bulbs, by the fundamental rule of counting, there are 6*5*9 = 270 ways of selecting 3 bulbs with exactly two 75-W bulbs.

So, the probability of selecting exactly 2 bulbs of 75 W is

\frac{270}{455}=0.5934=59.34\%

(b)

The probability of selecting three 40-W bulbs is

\frac{4*3*2}{455}=0.0527=5.27\%

The probability of selecting three 60-W bulbs is

\frac{5*4*3}{455}=0.1318=13.18\%

The probability of selecting three 75-W bulbs is

\frac{6*5*4}{455}=0.2637=26.37\%

Since <em>the events are disjoint</em>, the probability of taking 3 bulbs of the same kind is the sum 0.0527+0.1318+0.2637 = 0.4482 = 44.82%

(c)

There are 6*5*4 ways of selecting one bulb of each type, so the probability of selecting 3 bulbs of each type is

\frac{6*5*4}{455}=0.2637=26.37\%

(d)

The probability that it is necessary to examine at least six bulbs until a 75-W bulb is found, <em>supposing there is no replacement</em>, is the same as the probability of taking 5 bulbs one after another without replacement and none of them is 75-W.

As there are 15 bulbs and 9 of them are not 75-W, the probability a non 75-W bulb is \frac{9}{15}=0.6

Since there are no replacement, the probability of taking a second non 75-W bulb is now \frac{8}{14}=0.5714

Following this procedure 5 times, we find the probabilities

\frac{9}{15},\frac{8}{14},\frac{7}{13},\frac{6}{12},\frac{5}{11}

which are

0.6, 0.5714, 0.5384, 0.5, 0.4545

As the events are independent, the probability of choosing 5 non 75-W bulbs is the product

0.6*0.5714*0.5384*0.5*0.4545 = 0.0419 = 4.19%

3 0
3 years ago
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Answer:

y=-30

Step-by-step explanation:

we are given

y varies directly as x

so, we can write equation as

y=kx

where

k is constant of proportionality

We have

at x=-1 , y=6

so, we can use it to find k

6=k(-1)

so, we get

k=-6

now, we can plug it back

y=-6x

now, we can plug x=5

and we get

y=-6\times 5

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Answer:

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Step-by-step explanation:

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