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Nesterboy [21]
3 years ago
7

Write the Roman numeral CLXXIX in standard form.​

Mathematics
2 answers:
defon3 years ago
6 0

179 is the answer

hope it helps

Gennadij [26K]3 years ago
3 0

Answer:

179 = CLXXIX = 100 + 50 + 10 + 10 + 10 − 1

Step-by-step explanation:

Numbers close to CLXXIX

Below are the numbers CLXXVI through CLXXXII, which are close to CLXXIX. The right column shows how each roman numeral adds up to the total.

176 = CLXXVI = 100 + 50 + 10 + 10 + 5 + 1

177 = CLXXVII = 100 + 50 + 10 + 10 + 5 + 1 + 1

178 = CLXXVIII = 100 + 50 + 10 + 10 + 5 + 1 + 1 + 1

179 = CLXXIX = 100 + 50 + 10 + 10 + 10 − 1

180 = CLXXX = 100 + 50 + 10 + 10 + 10

181 = CLXXXI = 100 + 50 + 10 + 10 + 10 + 1

182 = CLXXXII = 100 + 50 + 10 + 10 + 10 + 1 + 1

Symbol Value

I 1

V 5

X 10

L 50

C 100

D 500

M 1000

I hope this helps.

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Answer:

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3 years ago
The surface areas of two similar solids are 340 yd² and 1,158 yd². The volume of the larger solid is 1,712 yd³. What is the volu
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The ratio of surface area is equal to the ratio of the square of the corresponding dimensions. And ratio of volumes of two solids is equal to the cube of the ratio of the corresponding dimensions .

We start with the relation between ratio of surface area and ratio of corresponding sides. That is

\frac{340}{1158}=(\frac{x}{y} )^2

Here x and y are the corresponding sides .

\frac{x}{y} =\sqrt{\frac{340}{1158} } =0.542

Let the volume of the smaller one be v

\frac{v}{1712}=0.542^3

v=1712*0.542^3=272

So for the smaller solid, volume is 272 . And the correct option is the first option .

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3 years ago
The local oil changing business is very busy on Saturday mornings and is considering expanding. A national study of similar busi
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Answer:

t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131  

Rejection Zone: t_{calculated} or t_{calculated}>2.131

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts  and data given  

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".  

\bar X=4.2 represent the sample mean  

s=1.4 represent the sample standard deviation  

n=16 represent the sample selected  

\alpha=0.05 significance level  

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean is different from 3.6, the system of hypothesis would be:    

Null hypothesis:\mu = 3.6    

Alternative hypothesis:\mu \neq 3.6    

If we analyze the size for the sample is < 30 and we know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:    

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic  

We can replace in formula (1) the info given like this:    

t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714

Critical values

On this case since we have a bilateral test we need to critical values. We need to use the t distribution with df=n-1=16-1=15 degrees of freedom. The value for \alpha=0.05 and \alpha/2=0.025 so we need to find on the t distribution with 15 degrees of freedom two values that accumulates 0.025 of the ara on each tail. We can use the following excel codes:

"=T.INV(0.025,15)" "=T.INV(1-0.025,15)"

And we got t_{crit}=\pm 2.131    

So the decision on this case would be:

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131  

Rejection Zone: t_{calculated} or t_{calculated}>2.131

Conclusion    

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

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