Solve the equation 16y(2–y)+(4y–5)2=0

1 answer:

6 0

16y(2−y)+(4y−5)(2)=0

Step 1: Simplify both sides of the equation.

−16y2+40y−10=0

Step 2: Use quadratic formula with a=-16, b=40, c=-10.

y=−b±√b2−4ac2a

y=−(40)±√(40)2−4(−16)(−10)2(−16)

y=−40±√960−32

y=54+−14√15 or y=54+14√15

Answer:

y=54+−14√15 or y=54+14√15

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The Wall Street Journal reports that 33% of taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deduction

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Answer:

(a) n : 20 50 100 500

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

(b) The correct option is (b).

Step-by-step explanation:

Let the random variable X 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, μ = $16,642 and standard deviation is, σ = $2,400.

Assuming that the random variable X 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 n = 20

$P(\mu-200$

$=P(-0.37$

For a sample size of n = 50

$P(\mu-200$

$=P(-0.59$

For a sample size of n = 100

$P(\mu-200$

$=P(-0.83$

For a sample size of n = 500

$P(\mu-200$

$=P(-1.86$

n : 20 50 100 500

P (-200 < X - μ < 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 ( $\mu_{x}$ ).