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2 years ago

13

The Running Aces card team won $548 dollars playing in poker tournaments last year. The winnings were split evenly among the p p

layers.
How much money did each player receive?
Write your answer as an expression.

1 answer:

andriy [413]2 years ago

7 0

Since the question doesnt give the amount of players on the team we can just say that each player was given <u>548/p dollars.</u>

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Rob deposits $3,500 at Town Bank,

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3 years ago

Harvey collects rare marbles. he has 112 marbles in his collection. each tray holds an equal number of marbles. if he uses 7 tra

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3 years ago

Annual rent $ 7,380

jeyben [28]

Answer:

Part A:

Rent = $7380

Mortgage payments = $9800

Insurance = $145

Taxes, insurance, maintenance = $1780+1050$ = $2830

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3 years ago

A simple random sample of size n is drawn from a normally distributed population, and the mean of the sample is (x with a line o

prohojiy [21]

Answer:

Last option

${\overline {x}}} \± \frac{2.58*s}{\sqrt{n}$

Step-by-step explanation:

${\overline {x}}$

If we extract a sample of size n from a normal population and we want to estimate the population mean then the confidence interval for the mean will be:

${\overline {x}}} \± Z_{\alpha/ 2}*\frac{s}{\sqrt{n}$

Where

Where ${\overline {x}}$ is an estimator for the mean μ.

n is the sample size and s is its standard deviation.

$\alpha$ is the level of significance

$\alpha$ = 1-Confidence Level

$\alpha= 1-0.99$

$\alpha = 0.01$

Then $Z_{\frac{\alpha}{2}} = Z_{\frac{0.01}{2}} = Z_{0.005}=2.58$

Then the 99% confidence interval for the population mean is

${\overline {x}}} \± \frac{2.58*s}{\sqrt{n}$

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4 years ago

Read 2 more answers

For $x$, $y$, and $z$ positive real numbers, what is the maximum possible value for\[\sqrt{\frac{3x 4y}{6x 5y 4z}} \sqrt{\frac{y

joja [24]

Answer:

Infinity

Step-by-step explanation:

Since x,y,z are positive rel numbers, we have that

$\dfrac{3x4y}{6x5y4z}=\dfrac{1}{10z}\\\\\\\dfrac{y2z}{6x5y4z}=\dfrac{1}{60x}\\\\\\\dfrac{2z3x}{6x5y4z}=\dfrac{1}{20y}$

Hence,

$\sqrt{\dfrac{3x4y}{6x5y4z}}\sqrt{\dfrac{y2z}{6x5y4z}}\sqrt{\dfrac{2z3x}{6x5y4z}}\\\\\\=\sqrt{\dfrac{1}{10z}\dfrac{1}{60x}\dfrac{1}{20y}}=\sqrt{\dfrac{1}{12000xyz}}$

Now let

$f(x,y,z)=\sqrt{\dfrac{1}{12000 xyz}}$

if we take x=y=1, we have

$f(1,1,z)=\sqrt{\dfrac{1}{12000z}}$

and so f(1,1,z) tends to infinity as z goes to 0.

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4 years ago