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emmainna [20.7K]
3 years ago
11

Please answer fast will mark brainliest

Mathematics
2 answers:
bearhunter [10]3 years ago
7 0

Answer:

7/8

Step-by-step explanation:

Yes

tiny-mole [99]3 years ago
5 0

Answer:

option C is the correct one according to me

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A patient has 0.0000075 gram of iron in 1 liter of blood. The normal level is between 6* 10^-7 grams and 1.6*10^-5 grams. Is the
Vinil7 [7]
0.0000075 = 7.5 x 10^6 --- Patients Blood

and it has to be between 0.0000006 and 0.000016

so <em>Yes</em> it is normal
4 0
3 years ago
Se tiene que pegar encaje aun aro de 35 centimetros de diametro¿Cuanto encaje se necesita para cubrirlo?
nalin [4]
Cual es la formula?
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4 0
3 years ago
Read 2 more answers
Suppose 52% of the population has a college degree. If a random sample of size 563563 is selected, what is the probability that
amm1812

Answer:

The value is  P(| \^ p -  p| < 0.05 ) = 0.9822

Step-by-step explanation:

From the question we are told that

    The population proportion is  p =  0.52

     The sample size is  n  =  563      

Generally the population mean of the sampling distribution is mathematically  represented as

           \mu_{x} =  p =  0.52

Generally the standard deviation of the sampling distribution is mathematically  evaluated as

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

=>      \sigma  =  \sqrt{\frac{ 0.52 (1- 0.52 )}{563} }

=>      \sigma  =   0.02106

Generally the  probability that the proportion of persons with a college degree will differ from the population proportion by less than 5% is mathematically represented as

            P(| \^ p -  p| < 0.05 ) =  P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 ))

  Here  \^ p is the sample proportion  of persons with a college degree.

So

 P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P(\frac{[[0.05 -0.52]]- 0.52}{0.02106} < \frac{[\^p - p] - p}{\sigma }  < \frac{[[0.05 -0.52]] + 0.52}{0.02106} )

Here  

    \frac{[\^p - p] - p}{\sigma }  = Z (The\ standardized \  value \  of\  (\^ p - p))

=> P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P[\frac{-0.47 - 0.52}{0.02106 }  <  Z  < \frac{-0.47 + 0.52}{0.02106 }]

=> P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P[ -2.37 <  Z  < 2.37 ]

=>  P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P(Z <  2.37 ) - P(Z < -2.37 )

From the z-table  the probability of  (Z <  2.37 ) and  (Z < -2.37 ) is

  P(Z <  2.37 ) = 0.9911

and

  P(Z <  - 2.37 ) = 0.0089

So

=>P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) =0.9911-0.0089

=>P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = 0.9822

=> P(| \^ p -  p| < 0.05 ) = 0.9822

3 0
4 years ago
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 108.7-cm and a standard dev
Luden [163]

Let x be the lengths of the steel rods and X ~ N (108.7, 0.6)

To get the probability of less than 109.1 cm, the solution is computed by:


z (109.1) = (X-mean)/standard dev

= 109.1 – 108/ 0.6

= 1.1/0.6

=1.83333, look this up in the z table.


P(x < 109.1) = P(z < 1.8333) = 0.97 or 97%

3 0
3 years ago
Round 7.954 to the place named
Andrej [43]
What is it rounded too?
4 0
3 years ago
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