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Bezzdna [24]
1 year ago
6

Q: The ratio of the numbers of Nick's marbles to Michael's is 3:5. Nick has 36 marbles. If Nick receives

Mathematics
1 answer:
Nataly [62]1 year ago
7 0

Answer:

\frac{3}{4}

Step-by-step explanation:

\frac{3}{5}  =  \frac{36}{x}

3x = 36 x 5

3x = 180

x = 60

Michael has 60 marbles.

Nick recieves 9 more marbles so he has 36 + 9 = 45 marbles

The new ratio will be:

\frac{45}{60}  = \frac{3}{4}

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Answer: 4188.8

Step-by-step explanation:

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2 years ago
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Th correct answer would be D
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The length of each side of a regular pentagon is increased by 8 inches, so the perimeter is now 65 inches. What is the original
evablogger [386]

<em>Here</em> as the <em>Pentagon</em> is <em>regular</em> so it's <em>all sides</em> will be of <em>equal length</em> . And if we assume It's each side be<em> </em><em><u>s</u></em> , then it's perimeter is going to be <em>(s+s+s+s+s) = </em><em><u>5s</u></em>.And as here , each <em>side</em> is increased by <em>8 inches</em> and then it's perimeter is <em>65 inches</em> , so we got that it's side after increament is<em> (s+8) inches</em> and original length is <em>s inches </em>. And if it's each side is <em>(s+8) inches</em> , so it's perimeter will be <em>5(s+8)</em> and as it's equal to <em>65 inches</em> . So , <em><u>5(s+8) = 65</u></em>

{:\implies \quad \sf 5(s+8)=65}

{:\implies \quad \sf 5s+5\times 8=65}

{:\implies \quad \sf 5s+40=65}

{:\implies \quad \sf 5s=65-40}

{:\implies \quad \sf 5s=25}

{:\implies \quad \sf s=\dfrac{25}{5}=5}

{:\implies \quad \bf \therefore \quad \underline{\underline{s=5\:\: Inches}}}

As we assumed the original side to be <em><u>s</u></em> .

<em>Hence, the original side's length 5 inches </em>

8 0
2 years ago
In a certain assembly plant, three machines B1, B2, and B3, make 30%, 20%, and 50%, respectively. It is known from past experien
diamong [38]

Answer:

The probability that a randomly selected non-defective product is produced by machine B1 is 11.38%.

Step-by-step explanation:

Using Bayes' Theorem

P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|a)P(a)}

where

P(B|A) is probability of event B given event A

P(B|a) is probability of event B not given event A  

and P(A), P(B), and P(a) are the probabilities of events A,B, and event A not happening respectively.

For this problem,

Let P(B1) = Probability of machine B1 = 0.3

P(B2) = Probability of machine B2 = 0.2

P(B3) = Probability of machine B3 = 0.5

Let P(D) = Probability of a defective product

P(N) = Probability of a Non-defective product

P(D|B1) be probability of a defective product produced by machine 1 = 0.3 x 0.01 = 0.003

P(D|B2) be probability of a defective product produced by machine 2 = 0.2 x 0.03 = 0.006

P(D|B3) be probability of a defective product produced by machine 3 = 0.5 x 0.02 = 0.010

Likewise,

P(N|B1) be probability of a non-defective product produced by machine 1 = 1 - P(D|B1) = 1 - 0.003 = 0.997

P(N|B2) be probability of a non-defective product produced by machine 2  = 1 - P(D|B2) = 1 - 0.006 = 0.994

P(N|B3) be probability of a non-defective product produced by machine 3 = 1 - P(D|B3) = 1 - 0.010 = 0.990

For the probability of a finished product produced by machine B1 given it's non-defective; represented by P(B1|N)

P(B1|N) =\frac{P(N|B1)P(B1)}{P(N|B1)P(B1) + P(N|B2)P(B2) + (P(N|B3)P(B3)} = \frac{(0.297)(0.3)}{(0.297)(0.3) + (0.994)(0.2) + (0.990)(0.5)} = 0.1138

Hence the probability that a non-defective product is produced by machine B1 is 11.38%.

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<u>Answer: </u>

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We are given the following the quadratic function and we are to rewrite it in intercept or factored form:

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We can factorize the given function so taking the common factors out of it to get:

f(x)=3x^2 - 12

f(x) = 3 (x^2 - 4)

The term (x^2-4) is in the form a^2-b^2 so it can further be factorized to give:

f(x) = 3 (x+2)(x-2)

Therefore, the factored form of the given quadratic function is f(x) = 3(x+2)(x-2).


7 0
3 years ago
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