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viva [34]
2 years ago
11

Write the ratio as a fraction in simplest form. 51 correct : 9 incorrect

Mathematics
2 answers:
irinina [24]2 years ago
6 0
51/9 = 17/3. That is the answer. 
Aneli [31]2 years ago
4 0
I think it is 9/51= 3/17
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A rectangle has an area of 36 inches squared. Write an equation to show how the length (l) varies inversely the width (w).
Anastaziya [24]
Start with the first sentence, and the definition of the area of a rectangle:
l * w = 36

Rewrite the equation to get
l = 36 / w
or
w = 36 / l
Either one shows the inverse relationship between w and l.

Assuming the last part wants you to write an equation for an x-y graph, the equation would be
y = 36 / x
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A flag stick on a golf course that is 6 1/2 feet tall casts a 9 3/4 foot shadow while a golfer nearby casts a 8 3/4 foot shadow.
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Massa says that 540 = 6 = 9. Is he right? Explain how you know using an equation.
mixas84 [53]

Answer:

i think you have written wrong equation

Step-by-step explanation:

8 0
2 years ago
If mJK=(7x-39) and mML)=87, find x
lilavasa [31]
If two chords of a circle are congruent, then their intercepted arcs are congruent
7x-39 = 87
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5 0
2 years ago
The Wall Street Journal reports that 33% of taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deduction
Len [333]

Answer:

(a) <em>                             </em><em>n</em> :      20           50          100         500

P (-200 < <em>X</em> - <em>μ </em>< 200) : 0.2886    0.4444    0.5954    0.9376

(b) The correct option is (b).

Step-by-step explanation:

Let the random variable <em>X</em> represent the amount of deductions for taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return.

The mean amount of deductions is, <em>μ</em> = $16,642 and standard deviation is, <em>σ</em> = $2,400.

Assuming that the random variable <em>X </em>follows a normal distribution.

(a)

Compute the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean as follows:

  • For a sample size of <em>n</em> = 20

P(\mu-200

                                           =P(-0.37

  • For a sample size of <em>n</em> = 50

P(\mu-200

                                           =P(-0.59

  • For a sample size of <em>n</em> = 100

P(\mu-200

                                           =P(-0.83

  • For a sample size of <em>n</em> = 500

P(\mu-200

                                           =P(-1.86

<em>                                  n</em> :      20           50          100         500

P (-200 < <em>X</em> - <em>μ </em>< 200) : 0.2886    0.4444    0.5954    0.9376

(b)

The law of large numbers, in probability concept, states that as we increase the sample size, the mean of the sample (\bar x) approaches the whole population mean (\mu_{x}).

Consider the probabilities computed in part (a).

As the sample size increases from 20 to 500 the probability that the sample mean is within $200 of the population mean gets closer to 1.

So, a larger sample increases the probability that the sample mean will be within a specified distance of the population mean.

Thus, the correct option is (b).

8 0
2 years ago
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