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lisov135 [29]
2 years ago
5

A company makes a profit of $1.38 million. This is $2.54 million more than last year. Write and solve an equation to find the pr

ofit p (in millions of dollars) last year.
An equation is
=1.38. ??

The profit last year was $(??)
million.
Mathematics
2 answers:
Drupady [299]2 years ago
7 0
I like your cut G ?!!!!!!!!!!!!
Alexus [3.1K]2 years ago
5 0

Answer:

-1.16

Step-by-step explanation:

1.38 - 2.54 = p

solve that, and to explain it, it says that it was making 1.38 mil this year but last year it was 2.54 mil less so you would subtract 2.54 from 1.38 since 1.38 is the main answer, so last year they were in the negatives

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A rectangular swimming pool is to be built with an area of 1800 square feet. The owner wants 5-foot wide decks along either side
FrozenT [24]

The solution would be like this for this specific problem:

A = (l + 20) · (w + 10)

The area of the pool is 1800 square feet and can be computed also like l · w. Then l · w = 1800, and therefore l = 1800 w.

We plug in the values:

A = (1800 / w + 20) · (w + 10) = 1800 + 18000 / w + 20w + 200 = 18000 / w + 20w + 2000

dA /dw = − 18000 / w2 + 20

w = 30

l = 60

Since A = (l + 20) · (w + 10):

A = (60 + 20) · (30+ 10)

A = 3,200

<span>A(x) = 1800 + 2*10(x+10)+ 2*5(1800/x) </span><span>
<span>= 1800 + 20x + 200 + 18000/x </span>
<span>= 2000 + 20 x + 18000/x </span></span>

<span>20 - 18000/x^2 = 0 </span><span>
<span>x = sqrt (900) = 30 </span>
<span>Pool length dimension = 1800/x = 60 </span>
Lot dimensions: 80 x 40</span>

I am hoping that these answers have satisfied your queries and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.

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The volume of a prism is the product of its height and area of its base, V = Bh. A rectangular prism has a volume of 16y4 + 16y3
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The law of cosines is a^2+b^2-2abcosC=c^2. Find the value of 2abcosC.
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A random sample from a normal population is obtained, and the data are given below. Find a 90% confidence interval for . 114 157
melamori03 [73]

Answer:

103.160 \leq \sigma \leq 187.476

And the upper bound rounded to the nearest integer would be 187.

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 mean or variance 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.

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

Data given: 114 157 203 257 284 299 305 344 378 410 421 450 478 480 512 533 545

The confidence interval for the population variance \sigma^2 is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

On this case we need to find the sample standard deviation with the following formula:

s=sqrt{\frac{\sum_{i=1}^{17} (x_i -\bar x)^2}{n-1}}

And in order to find the sample mean we just need to use this formula:

\bar x =\frac{\sum_{i=1}^n x_i}{n}

The sample mean obtained on this case is \bar x= 362.941 and the deviation s=132.250

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=17-1=16

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a tabel to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.05,16)" "=CHISQ.INV(0.95,16)". so for this case the critical values are:

\chi^2_{\alpha/2}=26.296

\chi^2_{1- \alpha/2}=7.962

And replacing into the formula for the interval we got:

\frac{(16)(132.250)^2}{26.296} \leq \sigma \leq \frac{(16)(132.250)^2}{7.962}

10641.959 \leq \sigma^2 \leq 35147.074

Now we just take square root on both sides of the interval and we got:

103.160 \leq \sigma \leq 187.476

And the upper bound rounded to the nearest integer would be 187.

4 0
2 years ago
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