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igor_vitrenko [27]
3 years ago
6

What part of 3 12 is 1/2 ? I need this asap

Mathematics
1 answer:
Aliun [14]3 years ago
6 0
I assume you are asking one of three things, so I will answer all of them.

if you are asking what is 1/2 of 312, then multiply 312 by 1/2. 312 is really just 312/1, so
312 \times  \frac{1}{2}  =  \frac{312}{1}  \times  \frac{1}{2}
we multiply fractions by multiplying straight across
\frac{312 \times 1}{1 \times 2}  =  \frac{312}{2}  = 156


if you were asking what part of 3 1/2 is 1/2 then we have to recognize what 3 1/2 is as an improper fraction. to do this, we convert 3 into halves. how many halves of pizza are there in 3 whole pizzas? 6. so we have 6 halves and one half, which is a total of 7 halves.
\frac{6}{2}  +  \frac{1}{2}  =  \frac{7}{2}
now, we do the same thing as the first problem
\frac{7}{2}  \times  \frac{1}{2}  =  \frac{7 \times 1}{2 \times 2}  =  \frac{7}{4}

if you were asking what is half of 3.12, then we convert 1/2 to a decimal and multiply
\frac{1}{2}  = .5 \\ 3.12 \times .5 = 1.56
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Step-by-step explanation:

First we need to find the slope:

\frac{y_{2}-y_{1} }{x_{2}-x_{1} }

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Step-by-step explanation:

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The manager of a supermarket would like to determine the amount of time that customers wait in a check-out line. He randomly sel
garri49 [273]

Complete question:

The manager of a supermarket would like to determine the amount of time that customers wait in a check-out line. He randomly selects 45 customers and records the amount of time from the moment they stand in the back of a line until the moment the cashier scans their first item. He calculates the mean and standard deviation of this sample to be barx = 4.2 minutes and s = 2.0 minutes. If appropriate, find a 90% confidence interval for the true mean time (in minutes) that customers at this supermarket wait in a check-out line

Answer:

(3.699, 4.701)

Step-by-step explanation:

Given:

Sample size, n = 45

Sample mean, x' = 4.2

Standard deviation \sigma = 2.0

Required:

Find a 90% CI for true mean time

First find standard error using the formula:

S.E = \frac{\sigma}{\sqrt{n}}

= \frac{2}{\sqrt{45}}

= \frac{2}{6.7082}

SE = 0.298

Standard error = 0.298

Degrees of freedom, df = n - 1 = 45 - 1 = 44

To find t at 90% CI,df = 44:

Level of Significance α= 100% - 90% = 10% = 0.10

t_\alpha_/_2_, _d_f = t_0_._0_5_, _d_f_=_4_4 = 1.6802

Find margin of error using the formula:

M.E = S.E * t

M.E = 0.298 * 1.6802

M.E = 0.500938 ≈ 0.5009

Margin of error = 0.5009

Thus, 90% CI = sample mean ± Margin of error

Lower limit = 4.2 - 0.5009 = 3.699

Upper limit = 4.2 + 0.5009 = 4.7009 ≈ 4.701

Confidence Interval = (3.699, 4.701)

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3 years ago
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Answer:

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Answer:

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25 x 5 = 125

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2 years ago
Read 2 more answers
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