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Butoxors [25]
4 years ago
14

Least to greatest -1.65/2 -7/80.9 -6/5

Mathematics
1 answer:
bagirrra123 [75]4 years ago
3 0
-7,-6,-1.65,2,5,80.9 is least to greatest
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What is an equation in a slop-intercept from for the line that passes through the points (1,-3) and (3,1) ?
kenny6666 [7]
<span>m=(2y-y1)/(x2-x1) thats your answer</span>
3 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
Someone please help me!!!!
kondaur [170]

Answer:

9.5

Step-by-step explanation:

13x+1+12x-2+14x+1+109+132+128

collect all the like terms

13x+12x+14x

=39x

1+-2+1+109+132+128

=369

39x=369

divide both sides by 39

=9.5

8 0
3 years ago
The sequence an = 8, 13, 18, 23, ... is the same as the sequence a1 = 8, an = an-1 + 5.
harkovskaia [24]
Actually, this answer would be true. Why?

 The first equation is: a(sub <em>n</em>) = 8, 13, 18, 23

The second is: a(sub 1)=8 ; a(sub <em>n</em>)= a(sub <em>n</em>-1)+5
if you wish to find the second term, plug two into the equation for <em /><em>n</em> 
8+5=13
to find the third, plug the second term, 13, in for <em>n.</em> 
13+5=18.

Hope this helped! I know it's a bit on the late side, but at least you can get the general idea!
8 0
4 years ago
Read 2 more answers
I am a 2-digit number. The sum of my digits is 11. I am divisible by both 4 and 7
Alecsey [184]

Answer:

56 or 84

Step-by-step explanation:

8 0
3 years ago
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