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3 years ago

Solve for r. Show your work! 12 =R– (34 – 2) Helpppp!!

Mathematics

2 answers:

elena-s [515]3 years ago

6 0

Answer:

we are just going to start by using pemdas and making it easier to solve

Step-by-step explanation:

12=R-(34-2)

12= R-32

add 32 to both sides

32+12=R

so R=44

Hope this helps :)

amm18123 years ago

4 0

Answer:

r = 44

Step-by-step explanation:

12 = r - (34 - 2)

Solve for r

___________

To solve, we need to isolate r.

Distribute the negative sign with the numbers within the parenthesis :

12 = r - 34 + 2

Add like terms :

12 = r - 32

Add 32 to both sides :

44 = r

r = 44

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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.

Step-by-step explanation:

Let X = number of seconds in the batch.

The probability of the random variable X is, p = 0.31.

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X 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

For positive acute angles A and B, it is known that tan A = 35/12 and sin B = 20/29. Find the value of sin(A - B ) in the simple

almond37 [142]

Answer:

$\displaystyle \sin(A-B)=\frac{495}{1073}$

Step-by-step explanation:

We are given that:

$\displaystyle \tan(A)=\frac{35}{12}\text{ and } \sin(B)=\frac{20}{29}$

Where both A and B are positive acute angles.

And we want to find he value of sin(A-B).

Using the first ratio, we can conclude that the opposite side is 35 and the adjacent side is 12.

Then by the Pythagorean Theorem, the hypotenuse is:

$h = \sqrt{35^2 + 12^2} =37$

Using the second ratio, we can likewise conclude that the opposite side is 20 and the hypotenuse is 29.

Then by the Pythagorean Theorem, the adjacent is:

$a=\sqrt{29^2-20^2}=21$

Therefore, we can conclude that:

So, for A, the adjacent is 12, opposite is 35, and the hypotenuse is 37.

For B, the adjacent is 21, opposite is 20, and the hypotenuse is 29.

We can rewrite sin(A-B) as:

$\sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B)$

Using the above conclusions, this yields: (Note that since A and B are positive acute angles, all resulting ratios will be positive.)

$\displaystyle \sin(A-B)=\Big(\frac{35}{37}\Big)\Big(\frac{21}{29}\Big)-\Big(\frac{12}{37}\Big)\Big(\frac{20}{29}\Big)$

Evaluate:

$\displaystyle \sin(A-B)=\frac{735-240}{1073}=\frac{495}{1073}$

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Misha Larkins [42]

Answer:

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2 years ago

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trapecia [35]

Answer:

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Is a triangle with sides of 9in, 25in, 24in a right triangle

Veronika [31]

Try the case if the triangle was a right triangle.
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3 years ago