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3 years ago

A rectangular park is 85 yards wide and 120 yards long.

Mathematics

1 answer:

ki77a [65]3 years ago

3 0

Answer:

A=10506.25 Sq.yards

Step-by-step explanation:

Width of the park=85 yds

Lenght of the park=120 yds

Now,

A=L*W

A=120*85

A=10200 square yards. This is the area of your given rectangle.

Now,

P=2(L+W)

P=2(120+85)

P=2(205)

P=410 this is the perimeter of your given rectangle.

Now,

a=410/4=102.5

Now,calculate the area of newly given distance of each side:

A=a^2

A=102.5^2

A=10506.25 Sq. Yards

This is the new area and is larger than the area of the given rectangle which was 10200 square yards.

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The diagram below is divided into equal parts.which fraction of the parts is white?

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Answer:

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A meteorologist who sampled 4 thunderstorms found that the average speed at which they traveled across a certain state was 16 mi

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Answer:

The 90 % confidence interval for the mean population is (11.176 ; 20.824 )

Rounding to at least two decimal places would give 11.18 , 20.83

Step-by-step explanation:

Mean = x`= 16 miles per hour

standard deviation =s= 4.1 miles per hour

n= 4

$\frac{s}{\sqrt n}$ = 4.1/√4= 4.1/2= 2.05

1-α= 0.9

degrees of freedom =n-1= df= 3

∈ ( estimator t with 90 % and df= 3 from t - table ) 2.353

Using Students' t - test

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Putting values

16 ± 2.353 * 2.05

= 16 + 4.82365

20.824 ; 11.176

The 90 % confidence interval for the mean population is (11.176 ; 20.824 )

Rounding to at least two decimal places would give 11.18 , 20.83

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2 years ago

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In a survey of 3100 people who owned a certain type of car, 1240 said they would buy that type of car again. What percent of th

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40% of people would buy the car again.

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Step-by-step explanation:

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2 years ago

How many solutions does the equation have? -4x - 9 = 12 - 4x

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Answer:

No Solution

Step-by-step explanation:

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2 years ago

Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis

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Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = the interval of time between the eruption

So, X ~ Normal( $\mu=75, \sigma^{2} =20$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

P(X > 84 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{84-75}{20}$ ) = P(Z > 0.45) = 1 - P(Z $\leq$ 0.45)

= 1 - 0.6736 = 0.3264

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{13} } }$ ) = P(Z > 1.62) = 1 - P(Z $\leq$ 1.62)

= 1 - 0.9474 = 0.0526

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{20} } }$ ) = P(Z > 2.01) = 1 - P(Z $\leq$ 2.01)

= 1 - 0.9778 = 0.0222

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

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2 years ago