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nataly862011 [7]
3 years ago
15

Tell what whole number you can substitute for xx in the following list so the numbers are ordered from least to greatest. 1/x,x/

3,80%
Mathematics
2 answers:
Marysya12 [62]3 years ago
7 0

Answer:

Take x = 2.

Step-by-step explanation:

We have,

The numbers are \frac{1}{x},\frac{x}{3},0.8.

We need to find the value of x such that they are ordered from the least to the greatest.

Let us take, x = 2.

So, we get,

\frac{1}{x}=\frac{1}{2}=0.5

\frac{x}{3}=\frac{2}{3}=0.6

As, 0.5 < 0.6 < 0.8.

Thus, taking x = 2 gives the desired result.

olasank [31]3 years ago
4 0
X/3 has to be smaller tha 80%
so x can be 2 since 2/3=66% about
1/2 is less thn 2/3

answer is x is 2
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iragen [17]
Complementary equal 90 degrees. 
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Supplementary equal 180 degrees.
180-47=133 degrees

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For what value of g does the function f(g) = g2 + 3g equal 18?
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Answer:

B) 3

Step-by-step explanation:

f(3) = 3^2 + 3(3) = 9 + 9 = 18

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2x2 + 2x-4?
nexus9112 [7]
Answer:

2x^2 +2x-4
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2x^2-4x+2

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2(x^2+x-2)
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Write x as a difference

2(x^2x-x-2)
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Use a^2-2ab+b^2=(ab)^2

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Reduce the fraction with 2

x^2x-x-2
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Factor out x from the expression

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Factor out negative sign from the expression

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6 0
2 years ago
Which of the following orderd pairs is a solution of 3x+2y=-19<br>7,-1<br>6,-7<br>-6,-7<br>-7,1​
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A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as
Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.

Step-by-step explanation:

Let <em>X</em> = number of seconds in the batch.

The probability of the random variable <em>X</em> is, <em>p</em> = 0.31.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is:

P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

              =1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

                            ={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

8 0
3 years ago
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