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MrRa [10]
3 years ago
11

3. Three Bean Salad Puzzle #2

Mathematics
2 answers:
kirza4 [7]3 years ago
8 0

Answer: 1 lima bean, 2 red beans, 3 black-eyed peas

Step-by-step explanation:

If 1/6 of the beans are lima beans, then there are 6 * 1/6 lima beans in the salad. 6 * 1/6 = 6/6 = 1.

If 1/3 of the beans are red beans, then there are 6 * 1/3 red beans in the salad. 6 * 1/3 = 6/3 = 2.

If 1/2 of the beans are black-eyed peas, then there are 6 * 1/2 black-eyed peas in the salad. 6 * 1/2 = 6/2 = 3.

Slav-nsk [51]3 years ago
4 0

Answer:

Lima Beans: 6 x \frac{1}{6} = 1

Red Beans: \frac{1}{3} x 6 = 2

Black Eyed-Peas: \frac{1}{2} x 6 = 3

Hope this helps!

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3 years ago
A scientist claims that 7% 7 % of viruses are airborne. If the scientist is accurate, what is the probability that the proportio
Reptile [31]

Answer:

The probability is P(|p-\^{p}| >  0.03)  =   0.0040

Step-by-step explanation:

From the question we are told that

   The population proportion is p =  0.07

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   The sample size is n = 600

Generally the standard deviation is mathematically represented as

     \sigma_p  =  \sqrt{\frac{p (1 -p)}{n} }

=>    \sigma_p  =  \sqrt{\frac{0.07(1 -0.07)}{600} }    

=>    \sigma_p  =  0.010416    

Generally the probability that the proportion of airborne viruses in a sample of 600 viruses would differ from the population proportion by greater than 3% is mathematically represented as

      P(|p-\^{p}| >  0.03) =  1 - P(|p -\^{p}| \le 0.03)

=>   P(|p-\^{p}| >  0.03)  =  1 -  P(-0.03 \le p -\^{p} \le 0.03 )

Now  add p  to  both side of the inequality

=>   P(|p-\^{p}| >  0.03)  =  1 -  P( 0.07-0.03  \le \^{p} \le 0.03+ 0.07 )

=>   P(|p-\^{p}| >  0.03)  =  1 -  P(0.04 \le \^{p} \le 0.10 )

Now  converting the probabilities to their respective standardized score

=> P(|p-\^{p}| >  0.03)  =  1 -  P(\frac{0.04 - 0.07}{0.010416}  \le Z \le \frac{0.10 -0.07}{0.010416}  )

=> P(|p-\^{p}| >  0.03)  =  1 -  P(-2.88  \le Z \le 2.88 )

=>   P(|p-\^{p}| >  0.03)  =   1 - [P(Z \le 2.88) - P(Z \le -2.88)]

From the z-table  

       P(Z \le 2.88)  =  0.9980

and

       P(Z \le -2.88)  = 0.0020

So

     P(|p-\^{p}| >  0.03)  =   1 - [0.9980 - 0.0020]

=>   P(|p-\^{p}| >  0.03)  =   0.0040

     

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3 years ago
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6 0
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