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3 years ago

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Each baseball team has nine players and each football team has eleven players. Six schools have both baseball and football teams

. Four schools have only a baseball team. How many players are there for all ten schools?

2 answers:

jolli1 [7]3 years ago

6 0

156 players in all ten schools

svetoff [14.1K]3 years ago

5 0

<span>Each baseball team has nine players and each football team has eleven players. Six schools have both baseball and football teams. Four schools have only a baseball team. How many players are there for all ten schools. The answer is 56 players. The six schools have 54 plus 66 players (baseball and football respectively ie 9x6 and 11x6) and the four schools have 9x4 =36 players so 54 + 66 +36= 156 players total.</span>

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Consider the graph of the line y = x – 4 and the point (−4, 2).

Hitman42 [59]

Answer:

a. The slope of a line parallel to the given line is 1

b. A point on the line parallel to the given line, passing through (−4, 2), is (1,7)

c. The slope of the line perpendicular to the given line is -1

d. A point on the line perpendicular to the given line, passing through (−4, 2), is (3,-5)

Step-by-step explanation:

The equation of the line in Slope-intercept form is:

$y=mx+b$

Where m is the slope and b is the y-intercept.

a. For the line $y = x - 4$

You can identify that:

$m=1$

By definition, two lines are parallel if they have the same slope. Then, the slope of a line parallel to the given line is:

$m=1$

b. The equation of the line in Point-slope form is:

$y -y_1 = m(x - x_1)$

Where m is the slope and ( $x_1,y_1$ ) is a point of the line.

Given the point (-4,2), substitute this point and the slope of the line into the equation:

$y -2 = (x +4)$

Give a value to "x", substitute it into this equation and solve for "y":

For $x=1$ :

$y -2 = (1 +4)$

$y= 5+2$

$y= 7$

Then, you get the point (1,7)

c. The slopes of perpendicular lines are negative reciprocals, then the slope of a line perpendicular to the given line is:

$m=-\frac{1}{1}\\\\m=-1$

d. Given the point (-4,2), substitute this point and the slope of the line into the equation:

$y -2 = -1(x +4)$

$y -2 = -(x +4)$

Give a value to "x", substitute it into this equation and solve for "y":

For $x=3$ :

$y -2 = -(3 +4)$

$y= -7+2$

$y= -5$

Then, you get the point (3,-5)

8 0

3 years ago

Assume that there is a 4% rate of disk drive failure in a year. a. If all your computer data is stored on a hard disk drive wit

kap26 [50]

Answer:

a) 99.84% probability that during a year, you can avoid catastrophe with at least one working drive

b) 99.999744% probability that during a year, you can avoid catastrophe with at least one working drive

Step-by-step explanation:

For each disk drive, there are only two possible outcomes. Either it works, or it does not. The disks are independent. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

4% rate of disk drive failure in a year.

This means that 96% work correctly, $p = 0.96$

a. If all your computer data is stored on a hard disk drive with a copy stored on a second hard disk drive, what is the probability that during a year, you can avoid catastrophe with at least one working drive?

This is $P(X \ geq 1)$ when $n = 2$

We know that either none of the disks work, or at least one does. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$ . So

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.96)^{0}.(0.04)^{2} = 0.0016$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0016 = 0.9984$

99.84% probability that during a year, you can avoid catastrophe with at least one working drive

b. If copies of all your computer data are stored on four independent hard disk drives, what is the probability that during a year, you can avoid catastrophe with at least one working drive?

This is $P(X \ geq 1)$ when $n = 4$

We know that either none of the disks work, or at least one does. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$ . So

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{4,0}.(0.96)^{0}.(0.04)^{4} = 0.00000256$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.00000256 = 0.99999744$

99.999744% probability that during a year, you can avoid catastrophe with at least one working drive

7 0

3 years ago

In the year 2005, a total of 750 fish were introduced into a manmade lake. The fish population was expected to grow at a rate of

tangare [24]

The population of the fish is given by P(t) = 750(1 + 0.083)^t; where t is the number of years after 2005.
Here, t = 2050 - 2005 = 45

Population in 2050 = 750(1.083)^45 = 750(36.16) = 27,123

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2 years ago

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