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Savatey [412]
2 years ago
12

Carry out the following division.​

Mathematics
2 answers:
kifflom [539]2 years ago
4 0

Answer:

1: 8a²

2: 9a²b²c²

Step-by-step explanation:

1: 48a³ ÷ 6a

48÷6 = 8

8a²

2: 72a³b⁴c⁵ ÷ 8ab²c³

72÷ 8 =9

9a²b²c²

Temka [501]2 years ago
4 0

Sry , I could only solve the first one

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Write the standard form of the equation of a circle with radius 3 and center (11, 3)
Talja [164]

Answer:

<u>x² - 22x + y² - 6y + 121 = 0</u>

Step-by-step explanation:

<u>General Form of Equation</u>

  • (x - h)² + (y - k)² = r²

<u>Solving</u>

  • (x - 11)² + (y - 3)² = (3)²
  • x² - 22x + 121 + y² - 6y + 9 = 9
  • <u>x² - 22x + y² - 6y + 121 = 0</u>
4 0
2 years ago
Read 2 more answers
Please help with the following question
Alex17521 [72]
P1=(0,0)=(x1,y1)→x1=0, y1=0
P2=(3,-2)=(x2,y2)→x2=3, y2=-2

Slope: m=(y2-y1)/(x2-x1)
m=(-2-0)/(3-0)
m=(-2)/3
m=-2/3

Point-slope equation:
y-y1=m(x-x1)
y-0=(-2/3)(x-0)
y=-(2/3)x

Answer: The equation of the line is: y=-(2/3)x
5 0
4 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
In january of one winter, 3/10 of a wood pile was used in a wood stove. In February, 2/5 of the wood pile was used. What part of
arsen [322]
2/5 is equivalent to 4/10 so at the end of February 4/10 was used up
And at the end of January, 3/10 was used
so 4/10 + 3/10 = 7/10

At the end of February, 7/10 of the wood was used
6 0
3 years ago
a candy factory has to order boxes for its candy canes if they plan to make 300,000 candy canes and each box can hold 12 candy c
natta225 [31]
25,000 boxes should be ordered

Hope this helps and have a good day :D
6 0
3 years ago
Read 2 more answers
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