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ddd [48]
4 years ago
7

Can someone teach me integers?

Mathematics
1 answer:
vivado [14]4 years ago
3 0
An integer is a whole number that can be positive, negative, or zero.

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is 0.75 less or greater than 7 over 8                                                                                           
Oksanka [162]
0.75 is 3/4 or 6/8, which is less than 7/8
4 0
4 years ago
Read 2 more answers
Please help with question below.
Nonamiya [84]

Answer:

Step-by-step explanation:

We have to find the points that are on the graph by finding the order pair (x, y) We are given the x value, so substitute x in the given equation f(x) and find y.

The graph is a line

6 0
3 years ago
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
Why are the solutions to the proportions StartFraction 40 over 8 EndFraction = StartFraction x over 10 EndFraction and StartFrac
UkoKoshka [18]
The answer is B x = 50
4 0
3 years ago
Read 2 more answers
What is the line of symmetry
Lerok [7]

Answer:

is the imaginary line where you could fold the image and have both halves match exactly.

6 0
3 years ago
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