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3 years ago

How do you express the area of the square as a monomial

Mathematics

1 answer:

Alenkasestr [34]3 years ago

4 0

If the length of the side of a square is ' Q ',
then the area of the square is Q² .

Q² is a monomial.

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The cash price of a television is $580.

Drupady [299]

Answer:

n=25% or 0.25

Step-by-step explanation:

865= 580(n%)+ 6(120)

865= 580 (n%) +720

865-720= 145

145= 580(n%)

divide both by 580, you get 0.25=n%

5 0

3 years ago

Please help me I honestly have zero clue how to answer this question

Iteru [2.4K]

Step-by-step explanation:

$n = 3 \times 5 {}^{2} \times {x}^{3}$

Square both sides of the equation.

${n}^{2} = (3 {}^{2} \times 5 {}^{4} \times {x}^{6} )$

$5 {n}^{2} = 5(3 {}^{2} \times {5}^{4} \times {x}^{6} )$

6 0

2 years ago

A sports tournament has d teams. Each team has 16 players . Using, write an expression for total number of players in the tourna

Vilka [71]

Answer:

16d

Step-by-step explanation:

multiply 16 by d as there are 16 players for each team

8 0

3 years ago

GRE Test: The Graduate Record Examination (GRE) is a test required for admission to many US graduate schools. Student’s scores o

Zepler [3.9K]

Answer:

a) $\bar X=144.3$

b) The 95% confidence interval would be given by (138.846;149.754)

c) n=34

d) n=298

e) n=1190

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".

We can calculate the sample mean with this formula:

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

$\bar X=144.3$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=8.8$ represent the population standard deviation

n=10 represent the sample size

a) Find a point estimate of the population mean

We can calculate the sample mean with this formula:

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

$\bar X=144.3$

b) Construct a 95% confidence interval for the true mean score for this population.

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $t_{\alpha/2}=1.96$

Now we have everything in order to replace into formula (1):

$144.3-1.96\frac{8.8}{\sqrt{10}}=138.846$

$144.3+1.96\frac{8.8}{\sqrt{10}}=149.754$

So on this case the 95% confidence interval would be given by (138.846;149.754)

Part c

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =+20 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$ , replacing into formula (5) we got: