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Vitek1552 [10]
3 years ago
8

There were 47 bales of hay in the barn. Mike stacked more bales in the barn today. There are now 78 of hay in the barn. How many

bales did he store in the barn?
Mathematics
2 answers:
ICE Princess25 [194]3 years ago
8 0

Answer:

31

Step-by-step explanation:

78 - 47 is 31 bales of hay

denis23 [38]3 years ago
5 0
He stored 31 more hay bales in the barn! How to find that out is you subtract 78 from 47 and that answer is 31 more added onto 47!

I hope this helped! Brainliest? :) -Raven❤️
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1. x= 3/Y as a fraction btw

2. Same thing
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Lena and Jade had a total of $36. Lena had twice as much as Jade. After each of them bought a movie ticket. Lena had 3 times as
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Is the relationship between the amount of gas and pressure direct or indirect? Explain your answer
guajiro [1.7K]

Answer:

INDIRECT!

see below

Step-by-step explanation:

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8 0
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Please help this is almost due soon!
PolarNik [594]

Answer:

6\pi

Step-by-step explanation:

The area of the circumference of a circle is:

C=2\pi r

We know that r is 8, so we can plug that into the formula.

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6 0
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

   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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