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spayn [35]
3 years ago
5

3[-x + (2 x + 1)] = x - 1 what is x

Mathematics
1 answer:
Nataly_w [17]3 years ago
4 0

The answer you are looking for is x=-2.

Solution/Explanation:

Writing out the equation

3[-x+(2x+1)]=x-1

Simplifying inside of the brackets first

Combining like terms, since -x+2x=x

3(x+1)=x-1

*You can remove the parenthesis, if preferred.

Using the Distributive Property on the left side of the equation

3x+3=x-1

Now, subtracting the "x" variable from both sides

3x+3-x=x-x-1

"x-x" cancels out to 0.

3x+3-x=-1

Combining like terms and simplifying

3x-x+3=-1

2x+3=-1

Subtracting 3 from both sides of the equation

2x+3-3=-1-3

"3-3" cancels out to zero.

2x+0=-1-3

2x=-1-3

Simplifying the right side of the equation

2x=-4

Finally, dividing both sides by 2

2x/2=-4/2

Simplifying the final part of the problem

Since 2x/2=x and -4/2=-2

x=-2

So, therefore, the final answer is x=-2.

Hope that this has helped you. Good day to you.

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Stanley Bank offers a money market account that earns 1.85% compounded continuously: a) If $10,000 is invested in this type of a
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Answer:

a) $1056.33 b) 23 years

Step-by-step explanation:

a) 10000(1+1.85/100)^3=10565.33 (2d.p.)

b) let x be the no. of years

15000 = 10000(1+1.85/100)^x

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3 years ago
Of 575 broiler chickens purchased from various kinds of food stores in different regions of a country and tested for types of ba
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Answer:

a) 0.68 - 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.630  

0.68 + 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.730  

And the 99% confidence interval would be given (0.630;0.730).  

b) We are 99% confident that the true proportion of infected chickens are between 0.63 (63%) and 0.73 (73%)

c) For this case the answer would be No. Since all the required assumptions in order to construct the confidence interval  are satisfied, so then the results can be assumed for all the population

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

p represent the real population proportion of interest  

\hat p =0.68 represent the estimated proportion

n=575 is the sample size required (variable of interest)  

z represent the critical value for the margin of error  

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

Part a

The confidence interval would be given by this formula  

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

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=2.58  

And replacing into the confidence interval formula we got:  

0.68 - 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.630  

0.68 + 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.730  

And the 99% confidence interval would be given (0.630;0.730).  

Part b

We are 99% confident that the true proportion of infected chickens are between 0.63 (63%) and 0.73 (73%)

Part c

For this case the answer would be No. Since all the required assumptions in order to construct the confidence interval  are satisfied, so then the results can be assumed for all the population

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