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Aleksandr [31]
1 year ago
5

Josslyn has nickels and dimes in her pocket. The number of nickels is three more than seven times the number of dimes let d repr

esent the number of dimes. Write the expression for the number of nickels
Mathematics
1 answer:
vfiekz [6]1 year ago
3 0
7d+3Explanation

to solve this we need to translate into math terms, so

Step 1

a) let d represents the number of dimes

let n represents the number of nickles

so

re write the expressions

\begin{gathered} number\text{ of dimes=d} \\ seven\text{ times the number of dimes = 7d} \\  \end{gathered}

The number of nickels is three more than seven times the number of dimes in other words you have to add 7 to seven times the number of dimes to obtain the number of nickles

hence

n=7d+3

therefore , the expression for the number of nickles is

7d+3

I hope this helps you

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Alborosie

Answer:

= \frac{34}{6561}

Step-by-step explanation:

= 3^{-8}\times \:34\\= 34\times \frac{1}{3^{8} } \\= \frac{1\times \:34}{3^{8} } \\= \frac{34}{3^{8} } \\= \frac{34}{6561}

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Answer:

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Step-by-step explanation:

y = -1/12x + b

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in a random sample of 200 people, 154 said that they watched educational television. fine the 90% confidence interval of the tru
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Answer:

[0.7210,0.8189] = [72.10%, 81.89%]

Step-by-step explanation:

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<em>We can approximate this discrete binomial distribution with the continuous Normal distribution. As the sample size is large enough, not applying the continuity correction factor makes no significant  difference. </em>

The approximation would be to a Normal curve with this parameters:

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The 90% confidence interval for the proportion would be then  

\bf  [0.77-z^*0.0298, 0.77+z^*0.0298]

where \bf z^* is the 10% critical value for the Normal N(0,1) distribution , this is a value such that the area under the N(0,1) curve outside the interval \bf [-z^*,z^*] is 10%=0.1

We can either use a table, a calculator or a spreadsheet to get this value.

In Excel or OpenOffice Calc we use the function

<em>NORMSINV(0.95) and we get a value of 1.645 </em>

The 90% confidence interval for the proportion is then

\bf  [0.77-1.645^*0.0298, 0.77+1.645^*0.0298]=[0.7210,0.8189]

This means there is a 90% probability that the proportion of people who watch educational television is between 72.10% and 81.89%

If the television company wanted to publicize the proportion of viewers, do you think it should use the 90% confidence interval?

Yes, I do.

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