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3 years ago

Will the following list have a large or small standard deviation?

Mathematics

2 answers:

Hunter-Best [27]3 years ago

7 0

Answer:

the standared deviation =2.87 so it is small

Step-by-step explanation:

to calculate the standard deviation of those numbers:

1- Work out the Mean (the simple average of the numbers)

(7+6+5+8+9+1+0+4+3+2)/10=4.5

Then for each number: subtract the Mean and square the result.

[(7 - 4.5)^2 + ... + (2 - 4.5)^2]/10=8.25

√8.25=2.87≅3

Sergio [31]3 years ago

5 0

It would be C large

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is your teacher trying to give you a stroke???

Step-by-step explanation:

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3 years ago

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In which of the following situations should the chi-square Test for Goodness-of-Fit be used? Select the correct answer below: a.

dangina [55]

Answer:

Correct answer is (a). An online research company puts out a survey asking people two questions. First, it asks whether they buy groceries online. Second, it asks whether they own a car or not. The data are collected and recorded in a contingency table. The company wants to determine if there is a relationship between buying groceries online and car ownership

Step-by-step explanation:

The chi-square distribution can be used to perform the goodness-of-fit test, Closeness of observed data to expected data of the model.

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3 years ago

Patricia deposited $612.40 in a savings account that earns 3.2% simple interest. What is Patricia's account balance after thirte

Tanzania [10]

Answer:

Patricia's account balance after thirteen years is $867.15.

Step-by-step explanation:

It is given that,

Principal on deposited amount is $612.40

Rate is 3.2%

We need to find Patricia's account balance after thirteen years, t = 13 years

This problem is based on the concept of simple interest. Interest is given by :

$I=PRt\\\\I=612.4\times \dfrac{3.2}{100}\times 13\\\\I=\$254.75$

Net amount = $612.40 + $254.75=$867.15

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4 years ago

train travelling at a uniform speed covers a distance of 255 km in Find the speed of the train and 3 4/5 1 1/2 the distance cove

serg [7]

The speed of the train is 150 km/hour.

The distance covered by the train in 3 4/5 hours is 570 km.

The speed of a body is calculated using the formula, speed = distance/time.

The distance covered is calculated using the formula, distance = speed*time.

The time taken by a body is calculated using the formula, time = distance/speed.

In the question, we are asked for the speed of a train, when it covers a distance of 225 km in 1 1/2 hours.

Distance = 225 km.

Time = 1 1/2 hours = 1.5 hours.

Speed = Distance/Time = 225/1.5 km/hour = 150 km/hour.

Now, we are asked to calculate the distance covered by the train in 3 4/5 hours.

Speed = 150 km/hour.

Time = 3 4/5 hours = 3.8 hours.

Distance = Speed*Time = 150*3.8 km = 570 km.

Thus, the speed of the train is 150 km/hour.

The distance covered by the train in 3 4/5 hours is 570 km.

The provided question is incorrect. The correct question is:

"A train at a uniform speed covers a distance of 225 km in 1 1/2 hours. Find the speed of the train and the distance covered in 3 4/5 hours.

Learn more about speed, time, and distance at

brainly.com/question/17146782

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2 years ago

Exclude leap years from the following calculations. (a) Compute the probability that a randomly selected person does not have a

Scrat [10]

Answer:

a) 99.73% probability that a randomly selected person does not have a birthday on March 14.

b) 96.71% probability that a randomly selected person does not have a birthday on the 2 nd day of a month.

c) 98.08% probability that a randomly selected person does not have a birthday on the 31 st day of a month.

d) 92.33% probability that a randomly selected person was not born in February.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

A non-leap year has 365 days.

(a) Compute the probability that a randomly selected person does not have a birthday on March 14.

There are 365-1 = 364 days that are not March 14. So

364/365 = 0.9973

99.73% probability that a randomly selected person does not have a birthday on March 14.

(b) Compute the probability that a randomly selected person does not have a birthday on the 2 nd day of a month.

There are 12 months, so there are 12 2nds of a month.

(365-12)/365 = 0.9671

96.71% probability that a randomly selected person does not have a birthday on the 2 nd day of a month.

(c) Compute the probability that a randomly selected person does not have a birthday on the 31 st day of a month.

The following months have 31 days: January, March, May, July, August, October, December.

So there are 7 31st days of a month during a year.

Then

(365-7)/365 = 0.9808

98.08% probability that a randomly selected person does not have a birthday on the 31 st day of a month.

(d) Compute the probability that a randomly selected person was not born in February.

During a non-leap year, February has 28 days. So

(365-28)/365 = 0.9233

92.33% probability that a randomly selected person was not born in February.

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3 years ago