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3 years ago

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What is the area of a piece of garden plot 5 yards 2 feet long by 6 yards 1 foot wide? Give your answer in square feet. A. 323 s

quare feet B. 270 square feet C. 70 square feet D. 49 square feet

1 answer:

Blababa [14]3 years ago

6 0

change yrds to ft

5yds = 5*3 = 15 ft

5yds 2 ft = 15ft + 2ft = 17 ft

6yds = 6 * 3 = 18 ft

6 yds 1 ft = 18 + 1

19 ft

A = l * w

A = 17* 19

A = 323 ft^2

Choice A

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A²- b² =<br> a² + b² =<br> (a + b)³ =

trasher [3.6K]

Answer:

a²- b²=(a+b)(a-b)

a² + b²=(a+b)²-2ab or (a-b)²+2ab

(a + b)³=a³+3a²b+3ab²+b³ or, a³+b³+3ab(a+b)

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Find the volume of the cylinder. Use 3.14 for T.<br><br> height of 5m radius of 15m

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Answer:

739.47

Step-by-step explanation:

height - h, radius -r

V=hπ²r=5×3.14²×15=739.47

5 0

3 years ago

Read 2 more answers

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

6. Jesse has 25 coins in his pocket, some are quarters, and some are dimes. If he has a

OLEGan [10]

Answer:

17 Quarters, 8 Dimes.

Step-by-step explanation:

17 x .25 = 4.25

8 x .10 = .80

4.25 + .80 = 5.05

5 0

3 years ago

suppose y varies directly with x, an dy = 10 when x =3. what direct variation equations relates x and y?

Ratling [72]

The direct variation equation relates x and y is y = $\frac{10}{3}$ x

Step-by-step explanation:

Direct variation is when one variable is equal to a constant times another variable

If y varies directly with x, then

y ∝ x
y = k x is the equation of variation
k is the constant of variation

∵ y varies directly with x

∵ y ∝ x

∴ y = k x

∵ y = 10 when x = 3

- Substitute these values in the equation above to find k

∴ 10 = k(3)

∴ 10 = 3 k

- Divide both sides by 3

∴ k = $\frac{10}{3}$

∴ The equation of variation is y = $\frac{10}{3}$ x

The direct variation equation relates x and y is y = $\frac{10}{3}$ x

Learn more:

You can learn more about variation in brainly.com/question/10708697

#LearnwithBrainly

8 0

3 years ago