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2 years ago

Calculate the area of each circle

Mathematics

1 answer:

Marysya12 [62]2 years ago

8 0

Answer:

$2.01 \: {inches}^{2}$

Step-by-step explanation:

The formula to find the area of a circle is

$area \: of \: circle \: = \pi {r}^{2}$

Let pi be (3.14), the radios as given in the question is (1.6 ÷ 2 = 0.8). Therefore;

$\pi {r}^{2} = \\ (3.14) \: {(0.8)}^{2} \\ 3.14 \times 0.64 = \\ 2.0096\: or \: 2.01{inches}^{2}$

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Write an equation for the following: y varies directly with x and z. Find K when x=2, y=12 and z=5.

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$\qquad \qquad \textit{direct joint variation} \\\\ \textit{\underline{y} varies directly with \underline{x} and \underline{z}}~\hspace{8.5em} \stackrel{\textit{constant of variation}}{y=\stackrel{\downarrow }{k}xz~\hfill } \\\\ \textit{\underline{y} varies directly with \underline{x}, }w^2\textit{ and }z^5~\hspace{6em} \stackrel{\textit{constant of variation}}{y=\stackrel{\downarrow }{k}xw^2z^5~\hfill } \\\\[-0.35em] ~\dotfill$

$y = kxz\qquad \textit{we know that} \begin{cases} x=2\\ y=12\\ z=5 \end{cases}\implies 12=k(2)(5) \\\\\\ \cfrac{12}{(2)(5)}=k\implies \cfrac{6}{5}=k~\hspace{10em}\boxed{y=\cfrac{6}{5}xz}$

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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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