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eduard
3 years ago
6

PLEASE HELP ILL GIVE U BRAINIEST❤️❤️

Mathematics
2 answers:
djverab [1.8K]3 years ago
7 0

Answer:

8 4/10, or if you simplify it, 8 2/5

Step-by-step explanation:

The mother cat weighs 14 times more than the kitten, which is 3/5 of a pound, so you multiply 3/5 by 14, to get 8 4/10, or 8 2/5.

horrorfan [7]3 years ago
7 0

Answer and Step-by-step explanation:

The mother cat weighs 8.4 pounds but as a fraction, this would be 8 2/5 (8 and 2 over 5) or 45/5. The mother cat weighs 8.4 pounds or 8 2/5 pounds.  You can tell this because whenever you multiply 3/5 x 14 you get 8.4 then you take 8.4 and turn it into a fraction and simplify.

<em>Hope you found this helpful! Have a wonderful day! </em>

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4 years ago
How much does the cherry vanilla ice cream cost per quart? (1.75 quarts costs $4)
alina1380 [7]

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Step-by-step explanation:

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3 years ago
Read 2 more answers
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

   The mean of the sampling distribution is   \mu_p =  0.07

   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

     

7 0
3 years ago
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Vadim26 [7]

Answer:

43.98

Step-by-step explanation:

did some quick maths

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c = 2 \times \pi \times r

8 0
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melisa1 [442]

Answer:point c

Step-by-step explanation:

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3 years ago
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