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Rus_ich [418]
2 years ago
8

8 1/2 - 6 1/4 please subtract

Mathematics
2 answers:
love history [14]2 years ago
6 0

Answer:

2.25 or 9/4 is the answer

Hope this helps

Alexandra [31]2 years ago
4 0

Answer:

2.25

Step-by-step explanation:

8.5-6.25

(8-6)+(0.5-0.25)

2+0.25

2.25

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16x^2-40x+25 Yes I know my phone is messed up but at least you got the answer

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What are the x-intercepts of x^2– 3x = 4?
Sindrei [870]

Answer:

x^2-3x=4

x^2-3x-4=0

x^2-4x+x-4=0

x(x-4)+1(x-4)=0

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Step-by-step explanation:

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3 years ago
The mean preparation fee H&R Block charged retail customers in 2012 was $183 (The Wall Street Journal, March 7, 2012). Use t
astraxan [27]

Answer:

a)0.6192

b)0.7422

c)0.8904

d)at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.

Step-by-step explanation:

Let z(p) be the z-statistic of the probability that the mean price for a sample is within the margin of error. Then

z(p)=\frac{ME*\sqrt{N}}{s } where

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  • s is the standard deviation of the population
  • N is the sample size

a.

z(p)=\frac{8*\sqrt{30}}{50 } ≈ 0.8764

by looking z-table corresponding p value is 1-0.3808=0.6192

b.

z(p)=\frac{8*\sqrt{50}}{50 } ≈ 1.1314

by looking z-table corresponding p value is 1-0.2578=0.7422

c.

z(p)=\frac{8*\sqrt{100}}{50 } ≈ 1.6

by looking z-table corresponding p value is 1-0.1096=0.8904

d.

Minimum required sample size for 0.95 probability is

N≥(\frac{z*s}{ME} )^2 where

  • N is the sample size
  • z is the corresponding z-score in 95% probability (1.96)
  • s is the standard deviation (50)
  • ME is the margin of error (8)

then N≥(\frac{1.96*50}{8} )^2 ≈150.6

Thus at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.

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