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2 years ago

8/15 multiplied by 5/6

2 answers:

5 0

To multiply two fractions, we multiply the top number (numerators) together and the bottom numbers together (denominators) and then simplify.

8 x 5 = 40
15 x 6 = 90

40/90 can be simplified (÷10) to 4/9

andriy [413]2 years ago

5 0

8/15×5/6
=8/18
=4/9
=0.444

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2 years ago

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nasty-shy [4]

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Step-by-step explanation:

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2 years ago

Find the GCF of the monomials: 32x^2 and 24x^2y

aev [14]

Answer:

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Explanation:

First, do prime factorization of each of the coefficients:

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24 ⇒ 2^3, 3

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1 year ago

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Helen [10]

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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

Me is the margin of error from the mean

s is the standard deviation of the population

N is the sample size

z(p)= $\frac{8*\sqrt{30}}{50 }$ ≈ 0.8764

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

z(p)= $\frac{8*\sqrt{50}}{50 }$ ≈ 1.1314

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

z(p)= $\frac{8*\sqrt{100}}{50 }$ ≈ 1.6

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

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.

7 0

3 years ago