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3 years ago

Help please I need help with this asap

Mathematics

1 answer:

solniwko [45]3 years ago

4 0

Answer:

The answer is A.

Step-by-step explanation:

Hope fully this helps.

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Based on historical data, your manager believes that 41% of the company's orders come from first-time customers. A random sample

TEA [102]

Answer:

The probability that the sample proportion is between 0.35 and 0.5 is 0.7895

Step-by-step explanation:

To calculate the probability that the sample proportion is between 0.35 and 0.5 we need to know the z-scores of the sample proportions 0.35 and 0.5.

z-score of the sample proportion is calculated as

z= $\frac{p(s)-p}{\sqrt{\frac{p*(1-p)}{N} } }$ where

p(s) is the sample proportion of first time customers
p is the proportion of first time customers based on historical data

N is the sample size

For the sample proportion 0.35:

z(0.35)= $\frac{0,35-0.41}{\sqrt{\frac{0.41*0.59}{72} } }$ ≈ -1.035

For the sample proportion 0.5:

z(0.5)= $\frac{0,5-0.41}{\sqrt{\frac{0.41*0.59}{72} } }$ ≈ 1.553

The probabilities for z of being smaller than these z-scores are:

P(z<z(0.35))= 0.1503

P(z<z(0.5))= 0.9398

Then the probability that the sample proportion is between 0.35 and 0.5 is

P(z(0.35)<z<z(0.5))= 0.9398 - 0.1503 =0.7895

6 0

3 years ago

The following data set represents the age of

Julli [10]

$\huge\text{Hey there!}$

$\huge\textsf{Breaking the meaning down in simpler terms}$
$\huge\textbf{Median:}\\\large\text{is understood to be \bf the number in the center (also known as}\\\large\textbf{the middle number)}$

$\huge\textbf{Mean:}\\\large\text{is understood to be the \bf total.}$

$\huge\textsf{Formulas for each term}$

$\huge\text{Median:}\\\large\textsf{To find the median you have to put each of your numbers in your data}\\\large\textsf{set from LEAST (smallest) to GREATEST (biggest). }$

$\huge\text{Mean:}\\\mathsf{\dfrac{sum\ of\ all\ terms}{number\ of\ terms}= mean}$

$\huge\textbf{SOLVING FOR THE QUESTION}$

$\huge\textsf{Median:}\\\\\large\textbf{Original data set: }\large\text{27, 52, 64, 41, 33, 38, 42, 60, 72, 68}\\\\\large\textbf{Conversion: }\large\text{27, 33, 38, 41, 42, 52, 64, 68, 72}\\\\\large\textsf{Make sure it is even between both sides of the data plot.}\\\\\large\text{It seems to even on BOTH sides of \boxed{\textsf{14}} so it could possibly be your}\\\large\text{median.}$

$\huge\textsf{Mean:}\\\\\large\textbf{Original data set: }\large\text{27, 52, 64, 41, 33, 38, 42, 60, 72, 68}\\\\\large\textsf{Your equation: }\mathsf{\dfrac{27 + 52 + 64 + 41 + 33 + 38 + 42 + 60 +72 + 68}{10}}\\\\\mathsf{\dfrac{79 + 64 + 41 + 33 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{143 + 41 + 33 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{184 + 33 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{217 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{255 + 42 + 60 + 72 + 68}{10}}$

$\mathsf{\dfrac{297 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{357 + 72 + 68}{10}}\\\\\mathsf{\dfrac{429 + 68}{10}}\\\\\mathsf{\dfrac{497}{10}}\\\\\mathsf{\approx 49.70}\\\\\large\text{From the looks of it \boxed{\rm{\dfrac{497}{10}}}\ or \boxed{\text{49.70}} could possibly be your mean.}$