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3 years ago

In a study of

1 answer:

5 0

We know that:
Mean = 82 mm and SD = 10 mm ( standard deviation )
82 - 3 * SD = 82 - 3 * 10 = 82 - 30 = 52 mm
82 + 3 * SD = 82 + 3 * 10 = 82 + 30 = 112 mm
Population between 52 and 112 mm is within +/- 3 standard deviations from the mean.
By the 66- 95 - 99.7 % rule it is: 99.7% of the test group.
0.977 * 500 = 498.5
Answer:
99.7 % of the test group have a diastolic pressure between 52 and 112 mm, or 498 men.

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The answer to 0.92 into a percentage

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Answer:

92%

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3 years ago

Write an equation for an ellipse centered at the origin, which has foci at (0,\pm\sqrt{63})(0,± 63 )left parenthesis, 0, comma

steposvetlana [31]

Answer:

$\frac{x^{2} }{4312 } + \frac{y^{2} }{8281 }$

Step-by-step explanation:

Since the foci are at(0,±c) = (0,±63) and vertices (0,±a) = (0,±91), the major axis is the y- axis. So, we have the equation in the form (with center at the origin) $\frac{x^{2} }{b^{2} } + \frac{y^{2} }{a^{2} }$ .

We find the co-vertices b from b = ±√(a² - c²) where a = 91 and c = 63

b = ±√(a² - c²)

= ±√(91² - 63²)

= ±√(8281 - 3969)

= ±√4312

= ±14√22

So the equation is

$\frac{x^{2} }{(14\sqrt{22}) ^{2} } + \frac{y^{2} }{91^{2} } = \frac{x^{2} }{4312 } + \frac{y^{2} }{8281 }$

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3 years ago

Consider a game in which a participant pays $2 to roll a die. The participant receives $3 if they roll a 1 (i.E. They go up by a

Sauron [17]

Answer:

The expected monetary value of a single roll is $1.17.

Step-by-step explanation:

The sample space of rolling a die is:

S = {1, 2, 3, 4, 5 and 6}

The probability of rolling any of the six numbers is same, i.e.

P (1) = P (2) = P (3) = P (4) = P (5) = P (6) = $\frac{1}{6}$

The expected pay for rolling the numbers are as follows:

E (X = 1) = $3

E (X = 2) = $0

E (X = 3) = $0

E (X = 4) = $0

E (X = 5) = $0

E (X = 6) = $4

The expected value of an experiment is:

$E(X)=\sum x\cdot P(X=x)$

Compute the expected monetary value of a single roll as follows:

$E(X)=\sum x\cdot P(X=x)\\=[E(X=1)\times \frac{1}{6}]+[E(X=2)\times \frac{1}{6}]+[E(X=3)\times \frac{1}{6}]\\+[E(X=4)\times \frac{1}{6}]+[E(X=5)\times \frac{1}{6}]+[E(X=6)\times \frac{1}{6}]\\=[3\times \frac{1}{6}]+[0\times \frac{1}{6}]+[0\times \frac{1}{6}]\\+[0\times \frac{1}{6}]+[0\times \frac{1}{6}]+[4\times \frac{1}{6}]\\=1.17$

Thus, the expected monetary value of a single roll is $1.17.

7 0

3 years ago