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Bingel [31]
2 years ago
11

Find the area and perimeter.

Mathematics
1 answer:
____ [38]2 years ago
7 0

Answer:

i have this question too and i cant figure it out.

Step-by-step explanation:

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soldi70 [24.7K]
I believe a would be on the first line.
8 0
2 years ago
Find the measure of the arc.<br><br> 42 degrees<br> 84 degrees<br> 21 degrees<br> 81 degrees
MA_775_DIABLO [31]

honestly I needed help with this too

7 0
2 years ago
Read 2 more answers
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
HELP ME!!! ITS DUE TOMMOROW!! Find the largest value of $r$ such that $6r^2+7r+8=6$.
Alex787 [66]
First you need to see that this is a quadratic. I need.to put all of the values on one side of equation to see what I got.
6r^2 + 7r + 8 = 6
6r^2 + 7r + 2 = 0

Now this one is difficult to factor so i will use quadratic equation:
[-b (+-) sqrt (b^2 - 4ac)] / (2a)

we know that a b and c are in a quadratic at these positions.
ax^2 + bx + c

so
[-7 (+-) sqrt (7^2 -(4)(6)(2)] / (2) (6)
[-7 (+-) sqrt (49 - 48)] / 12
[-7 (+-) 1] /12

split into the + and - for 2 answers
(-7 + 1) / 12
-6/12
-1/2

And
(-7 -1) /12
-8 / 12
-2/3

those are.the 2 answers

But but it says largest so -1/2
4 0
3 years ago
Read 2 more answers
Find x. Round to the nearest degree.
jeka94

The Law of Cosines features the 3 side lengths of a triangle, plus the measure of the angle opposite one of those sides.


We want angle x, which is opposite the side of length 39.


Then: a^2 = b^2 - 2ab cos C becomes 39^2 = 36^2 + 59^2 - 2(36)(59)cos x


or 1521 = 3481 + 1296 - 2(36)(59) cos x


Subtract (3481+1296) from both sides: 1521 - 4777 = -4248cos x

-3256 = -4248cos x

-3256

Then: cosx = --------------- = 0.766

-4248


Solving for x: x = arccos -0.766 = 0.698 radian, or 40 degrees (answer)

5 0
3 years ago
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