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svlad2 [7]
3 years ago
8

If 9x+2y2−3z2=132 and 9y−2y2+3z2=867, then x+y =

Mathematics
1 answer:
Phantasy [73]3 years ago
7 0
Looking at these two equations, you need to cancel out the variables with powers and the z. This can be simply done by adding them.

9y-2y^2+3z^2=867
+
9x+2y^2-3z^2=132
-----------------------------
9y+9x=999

Factor the 9 out of the left side.
9(x+y)=999
x+y=111

The answer is C (111).

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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
Find the value of N in the question
iris [78.8K]
N = 1/3 -0
n = 1/3

answer
n = 1/3
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Three percent of the caterpillars metamorphosed into butterflies. If Ramona could count 120 butterflies, how many caterpillars h
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The answer to your problem would be 400 because to find the answer you need to multiply 120 by .3 to get the answer because 3 % means per 100 and that is why you make it .3 hope i help.
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3 years ago
Evaluate -1/3-(-4/9)​
bija089 [108]

Answer:

To subtract fractions, find the LCD and then combine.

Exact Form:

1/9

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0.¯1

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Y=3x+4 and crosses through the point (1,8)
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