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Katyanochek1 [597]

2 years ago

12

Chapter 7 of the Jiuzhang suanshu presents a problem of two linear equations involving acres of land and their respective prices

. One of the two equations can be translated to: 300x + y = 10000 If y = 87.5, what is the value for x? 300x + y = 10,000 300x + (87.5) = 10,000 300x + 6,250 = 10,000 x =

2 answers:

djyliett [7]2 years ago

3 1

Answer: 33.041666

Step-by-step explanation:

Given: The translated equation : $300x+y = 10000$

To find the value of x if y = 87.5, we need to substitute the value of y in the given translated equation, we will get

$300x+87.5 = 10000$

Subtract 87.5 from both the sides , we get

$300x=9912.5$

Now, divide 300 on both the sides, we get

$x=\frac{9912.5}{300}\\\\\Rightarrow x=33.041666666$

Rama09 [41]2 years ago

4 0

Answer: 12.5

Step-by-step explanation: I got u guys :)

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Answer:

$h^{*} = 0.23\,m$

Step-by-step explanation:

The scale factor described denotes a reduction operation. The missing dimension is given by simple rule of three:

$h^{*} = \frac{1}{8}\cdot (1.84\,m)$

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Evaluate. 58−(14)2=58-142= ________

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For this case we have the following expression:

$58- (14) ^ 2$

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$58- (14) ^ 2 = 58-196$

Then, the second step is to subtract both resulting numbers:

$58- (14) ^ 2 = -138$

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Answer:

The result of the expression is given by:

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The Bakhshali Manuscript shows that another method for calculating nonperfect squares was being used in India by 400 CE. Use thi

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2 years ago

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During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of str

Brilliant_brown [7]

Answer:

number of successes

$k = 235$

number of failure

$y = 265$

The criteria are met

A

The sample proportion is $\r p = 0.47$

B

$E =4.4 \%$

C

What this mean is that for N number of times the survey is carried out that the which sample proportion obtain will differ from the true population proportion will not more than 4.4%

Ci

$r = 0.514 = 51.4 \%$

$v = 0.426 = 42.6 \%$

D

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

E

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

F

Yes our result would support the claim because

$\frac{1}{3 } \ of N < \frac{1}{2} (50\%) \ of \ N , \ Where\ N \ is \ the \ population\ size$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 500$

The sample proportion is $\r p = 0.47$

Generally the number of successes is mathematical represented as

$k = n * \r p$

substituting values

$k = 500 * 0.47$

$k = 235$

Generally the number of failure is mathematical represented as

$y = n * (1 -\r p )$

substituting values

$y = 500 * (1 - 0.47 )$

$y = 265$

for approximate normality for a confidence interval criteria to be satisfied

$np > 5 \ and \ n(1- p ) \ >5$

Given that the above is true for this survey then we can say that the criteria are met

Given that the confidence level is 95% then the level of confidence is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha =0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{\r p (1- \r p}{n} }$

substituting values

$E = 1.96 * \sqrt{ \frac{0.47 (1- 0.47}{500} }$

$E = 0.044$

=> $E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the proportion obtain will differ from the true population proportion of those that are happy by more than 4.4%

The 95% confidence interval is mathematically represented as

$\r p - E < p < \r p + E$

substituting values

$0.47 - 0.044 < p < 0.47 + 0.044$

$0.426 < p < 0.514$

The upper limit of the 95% confidence interval is $r = 0.514 = 51.4 \%$

The lower limit of the 95% confidence interval is $v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } < \frac{1}{2} (50\%)$

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2 years ago

Solve the equation. Negative StartFraction 9 Over 15 EndFraction y + StartFraction 3 Over 21 EndFraction = StartFraction 5 Over

mina [271]

Answer:

Step-by-step explanation:

9/15y + 3/21 = 5/15 y - 14/21. Adding 14/21 to both sides we then get 9/15y + 17/21 = 5/15y. Subtracting 5/15y from both sides, we get 4/15y + 17/21 = 0.

Then we get 4/15y = -17/21. Dividing by 4/15 gets us -17/21 * 4/15 = -68/315 = y.

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2 years ago

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