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1 year ago

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The area of a rectangle is (27x5 + 9x4 - 18x3) square feet. The width of the rectangle is 9x2 feet. What is the length of the re

ctangle?
DO NOT INCLUDE PARENTHESIS IN YOUR ANSWER

2 answers:

Yuliya22 [10]1 year ago

8 0

If the width is 9x², then the length is 27x^5+9x^4-18x^3/9x², which equals 3x³+x²+2x. ☺☺☺☺

Usimov [2.4K]1 year ago

5 0

Answer:

The length of the rectangle is given by 3x^3+x^2-2x

Step-by-step explanation:

We have been given that

area of rectangle is $A=27x^5 + 9x^4 - 18x^3$

The width of the rectangle is $w=9x^2$

We know the area of rectangle is given by

$A=lw\\\\l=\frac{A}{w}$

On substituting the known values, we get

$l=\frac{27x^5 + 9x^4 - 18x^3}{9x^2}$

On simplifying this, we get

$l=\frac{27x^5}{9x^2} +\frac{9x^4}{9x^2} -\frac{18x^3}{9x^2} \\\\l=3x^3+x^2-2x$

Therefore, the length of the rectangle is given by 3x^3+x^2-2x

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A farmer plows a corn field in the shape of a rectangle that has an area of 15840 square yards if the length of the field is 132

igomit [66]

Answer:

<u>Width= 120</u>

<u />

Step-by-step explanation:

We know that <u>L*W= A</u>

So lets put in the variables we know into the equation.

<u>L= 132 yards</u>

<u>A= 15840</u>

132 * W= 15840

All we need to do is <u>divide 15840 by 132</u>

<u>15840/132= 120</u>

<u>W=120</u>

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2 years ago

If π is rounded to a whole number, what is a reasonable estimate for the circumference of a circle with a diameter of 9 cm?

mel-nik [20]

Pi rounded to a whole number would equal three. Circumference equals 2pi*r. The radius in this case is half of the diameter, so 4.5 cm. 2 times pi, or three equals 6 times 4.5 is 27

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2 years ago

Use right triangle DCB to find the trig values for angle D. sinD = cosD = tanD =

Sever21 [200]

Sin D=35/37
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Use SOHCAHTOA

4 0

2 years ago

Read 2 more answers

A newspaper vendor sells three papers to 135 customers, the papers are the daily times, the observer and the new Nigeria. 70 cus

IRINA_888 [86]

Answer:

3 customers brought all the three papers.

Step-by-step explanation:

Let, D, O and N represents the three newspapers.

D = The daily times

O = The observer

N = The new Nigeria.

According to the given information,

$n(D\cup O\cup N)=135, n(D)=70,n(O)=60,n(N)=50$

$n(D\cap O)=17, n(O\cap N)=15, n(D\cap N)=16$

We know that,

$n(A\cup B\cup C) = n(A) + n(B) + n(C)-n(A\cap B)-n(B\cap C)-n(C\cap A) + n(A\cap B\cap C)$

Using this formula, we get

$n(D\cup O\cup N) = n(D)+n(O) + n(N)-n(D\cap O)-n(O\cap N)-n(N\cap D) + n(D\cap O\cap N)$

$135 =70+60+50-17-15-16 +n(D\cap O\cap N)$

$135 =132+n(D\cap O\cap N)$

$135 -132=n(D\cap O\cap N)$

$3=n(D\cap O\cap N)$

Therefore, 3 customers brought all the three papers.

5 0

2 years ago

Hard times In June 2010, a random poll of 800 working men found that 9% had taken on a second job to help pay the bills. (www.ca

Phoenix [80]

Answer:

a) The 95% confidence interval would be given (0.0702;0.1098).

We can conclude that the true population proportion at 95% of confidence is between (0.0702;0.1098)

b) Since the confidence interval NOT contains the value 0.06 we have anough evidence to reject the claim at 5% of significance.

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

$\hat p =0.09$ represent the estimated proportion for the sample

n=800 is the sample size required

$z$ represent the critical value for the margin of error

Confidence =0.95 or 95%

Part a

The population proportion have the following distribution

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

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 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

The margin of error is given by :

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

$Me=1.96 \sqrt{\frac{0.09(1-0.09)}{800}}=0.0198$

And replacing into the confidence interval formula we got:

$0.09 - 1.96 \sqrt{\frac{0.09(1-0.09)}{800}}=0.0702$

$0.09 + 1.96 \sqrt{\frac{0.108(1-0.09)}{800}}=0.1098$

And the 95% confidence interval would be given (0.0702;0.1098).

We can conclude that the true population proportion at 95% of confidence is between (0.0702;0.1098)

Part b

Since the confidence interval NOT contains the value 0.06 we have anough evidence to reject the claim at 5% of significance.

6 0

1 year ago