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3 years ago

Graph the function defined by f(x)= |-x-7|

Mathematics

1 answer:

Charra [1.4K]3 years ago

7 0

Answer: The answer is the third graph, attached herewith the image.

Step-by-step explanation: The given function is

$f(x)=|-x-7|.$

We are to select the correct graph from the given four options that is defined by the function above.

At x = 0, y = 7

and

at y = -7, y = 0.

So, the points (0, 7) and (-7, 0) satisfies the given function and hence these points must lie on the graph also.

We can see that the points (0, 7) and (-7, 0) lie only on the third graph. The image is attached with these two points plotted.

Thus, the answer is the third graph.

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The population of a town is increasing by 300 inhabitants each year. If its population at the beginning of 1990 was 21,152, what

babunello [35]

Answer:

Step-by-step explanation:

You take the starting year of 1990 and subtract it from 1999 to get the year span of 9,

You then take the amount per year the population goes up by which is 300, so 300 multiplied by 9 is 2,700

You add 2,700 to 21,152 to get B-23,852

8 0

3 years ago

Maria mixes \frac{1}{4} cup white sugar and \frac{2}{3} cup brown sugar to make a topping for some muffins. She uses \frac{1}{12

Hoochie [10]

Answer: 11 muffins

Step-by-step explanation:

Since Maria mixes 1/4 cup white sugar with 2/3 cup brown sugar, this will give us a mixture of:

= 1/4 + 2/3

= 3/12 + 8/12

= 11/12

Ww are then told that she uses 1/12 cup of the mixture to top each muffin. Therefore, the number of muffins that she can top will be:

= 11/12 ÷ 1/12

= 11/12 × 12/1

= 12

She can top 11 muffins.

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2 years ago

What is the y intercept of the equation of the line that is perpendicular to the line y=3/5x+10 and passes through the point (15

Mama L [17]

It's 20/3. hope it will help you!

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2 years ago

What is 25 meters per second to miles per hour

leva [86]

Answer:

25 × 60 (1 minute) × 60 (1 hour) = 90,000 meters/hour

90,000 converted into miles is 55.923

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3 years ago

A highway traffic condition during a blizzard is hazardous. Suppose one traffic accident is expected to occur in each 60 miles o

o-na [289]

Answer:

a) 0.3408 = 34.08% probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway.

b) 0.0879 = 8.79% probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway.

c) 0.4493 = 44.93% probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway.

Step-by-step explanation:

We have the mean for a distance, which means that the Poisson distribution is used to solve this question. For item b, the binomial distribution is used, as for each blizzard day, the probability of an accident will be the same.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

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.

Suppose one traffic accident is expected to occur in each 60 miles of highway blizzard day.

This means that $\mu = \frac{n}{60}$ , in which 60 is the number of miles.

(a) What is the probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway?

$n = 25$ , and thus, $\mu = \frac{25}{60} = 0.4167$

This probability is:

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

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.4167}*(0.4167)^{0}}{(0)!} = 0.6592$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.6592 = 0.3408$

0.3408 = 34.08% probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway.

(b) Suppose there are six blizzard days this winter. What is the probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway?

Binomial distribution.

6 blizzard days means that $n = 6$

Each of these days, 0.6592 probability of no accident on this stretch, which means that $p = 0.6592$ .

This probability is P(X = 2). So

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

$P(X = 2) = C_{6,2}.(0.6592)^{2}.(0.3408)^{4} = 0.0879$

0.0879 = 8.79% probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway.

(c) If the probability of damage requiring an insurance claim per accident is 60%, what is the probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway?

Probability of damage requiring insurance claim per accident is of 60%, which means that the mean is now:

$\mu = 0.6\frac{n}{60} = 0.01n$

80 miles:

This means that $n = 80$ . So

$\mu = 0.01(80) = 0.8$

The probability of no damaging accidents is P(X = 0). So

$P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493$

0.4493 = 44.93% probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway.

6 0

2 years ago