Take away 4/5 from 6 1/2

Mathematics

2 answers:

Rashid [163]2 years ago

8 0

Hi there!

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I believe your answer is:

$5\frac{7}{10}$

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Here’s why:

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$\boxed{\text{Calculating the answer...}}\\\\6\frac{1}{2}-\frac{4}{5} \\-------------\\6\frac{1}{2} = \frac{13}{2} \\\\\frac{13}{2} - \frac{4}{5}\\\\LCM(2,5): 10\\\\\frac{13}{2} =\frac{13*5}{2*5} =\frac{65}{10}\\\\\frac{4}{5}=\frac{4*2}{5*2}=\frac{8}{10}\\\\\frac{65}{10}-\frac{8}{10}\\\\ \frac{57}{10}\\\\\frac{57}{10}\rightarrow\boxed{5\frac{7}{10}}$

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Hope this helps you. I apologize if it’s incorrect.

Dmitry [639]2 years ago

6 0

Answer:

3-4/5=2.2

hope it helps

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In a survey of 603 adults, 98 said that they regularly lie to people conducting surveys. Create a 99% confidence interval for th

marysya [2.9K]

Answer:

The population proportion is estimated to be with 99% confidence within the interval (0.1238, 0.2012).

Step-by-step explanation:

The formula for estimating the population proportion by a confidence interval is given by:

$\hat{p}\pm z_{\alpha /2}\times\sqrt{\frac{\hat{p}\times(1-\hat{p})}{n}}$

Where:

$\hat{p}$ is the sample's proportion of success, which in this case is the people that regularly lie during surveys,

$z_{\alpha /2}$ is the critical value needed to find the tails of distribution related to the confidence level,

$n$ is the sample's size.

<u>First</u> we compute the $\hat{p}$ value:

$\hat{p}=\frac{successes}{n}=\frac{98}{603}=0.1625$

<u>Next</u> we find the z-score at any z-distribution table or app (in this case i've used StatKey):

$z_{\alpha /2}=2.576$

Now we can replace in the formula with the obtained values to compute the confidence interval:

$\hat{p}\pm z_{\alpha /2}\times\sqrt{\frac{\hat{p}\times(1-\hat{p})}{n}}=0.1625\pm 2.576\times\sqrt{\frac{0.1625\times(1-0.1625)}{603}}=(0.1238, 0.2012)$