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4 years ago

In Pensacola, Florida, the average rainfall

Mathematics

1 answer:

Lerok [7]4 years ago

5 0

the average rainfall, in inches, in December is 4.15 inch . Correct option C. 4.15 .

<u>Step-by-step explanation:</u>

Here we have , In Pensacola, Florida, the average rainfall in the month of July is 7.40 inches. By October, the average rainfall decreases by 35%. By December, the average rainfall drops another 0.66 inches. We need to find the average rainfall, in inches, in December .Let's find out:

In July , rainfall was 7.40 inches . By October, the average rainfall decreases by 35% i.e.

⇒ $7.4 - \frac{7.4(35)}{100}$

⇒ $7.4 - 0.074(35)$

⇒ $7.4 - 2.59$

⇒ $4.81$

By December, the average rainfall drops another 0.66 inches i.e.

⇒ $4.81-0.66$

⇒ $4.15$

Therefore, the average rainfall, in inches, in December is 4.15 inch . Correct option C. 4.15 .

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1. Find the quartiles for the tire pressures of car tires at an auto clinic. Tire pressure is measured in psi

marusya05 [52]

The quartiles and outliers for the data provided are as follows:

1. Quartiles for the tire pressures of car tires at an auto clinic:

Q2 = 28 psi
Q4 = 36 psi
Q1 = 21.5 psi
Q3 = 34 psi

2. The outlier(s) for the tires prices is $450

3. The outlier(s) for the tires prices is $650

4. The outlier(s) for the tires prices is $540

5. Quartiles for the tire pressures of car tires at an auto clinic are:

Q2 = 27 psi
Q4 = 35 psi
Q1 = 23.5 psi
Q3 = 33 psi

<h3>Quartiles</h3>

Quartiles are the values that divide a list of numbers into four parts or quarters after the numbers are arranged in ascending order.

They describe the dispersionof a set of data values.

Q1 is the first quartile or lower quartile. 25% of the numbers in the data set are at or below Q1.
Q2 is the second quartile. 50% of the numbers are below Q2, and
Q3 is the third quartile, or upper quartile. 75% of the numbers are at or below Q3.
Q4 is the maximum value in the data set. 100% of the numbers are at or below Q4.

<h3>Outliers</h3>

Outliers are data points that differs significantly from other observations.

There are two outliers: lower boundary outliers and upper boundary outliers.

Formula for lower boundary outliers is:

Q1 − 1.5(IQR)

Formula for upper boundary outliers is:

Q3 + 1.5(IQR)

where IQR is interquartile range

IQR = Q3 - Q1

<h3>1. Quartiles for the tire pressures of car tires at an auto clinic: (20, 27,19, 23, 29, 28, 34, 34, 36)</h3>

Arranging in increasing order:

19, 20, 23, 27, 28, 29, 34, 34, 36.

Q2 = 28

Q4 = 36

Q1 = (20 + 23) / 2

Q1 = 21.5

Q3 = (34 + 34) /2

Q3 = 34

<h3>2. Outlier(s) for these tires prices: $58, $95, $78, $125, $87, $158, $152, $182, $195, $450</h3>

Arranging in increasing order:

58, 78, 87, 95, 125, 152, 158, 182, 195, 450

Q1 = (87 + 78)/2

Q1 = 62.5

Q3 = 182 + 195/2

Q3 = 188.5

IQR = 188.5 - 62.5

IQR = 126

Lower boundary outlier(s) = 62.5 - 1.5(126)

Lower boundary outliers = -126.5

Thus there are no lower outliers.

Upper boundary outliers = 188.5 + 1.5(126)

Upper boundary outliers = 377.5

Therefore, 450 is an outlier.

<h3>3. The outlier(s) for these tires prices: $78, $195, $98, $145, $87, $138, $159, $172, $155, $210, $240, $650</h3>

Arranging in increasing order:

78, 87, 98, 138, 145, 155, 159, 172, 195, 210, 240, 650

Q1 = 98

Q3 = 210

IQR = 210 - 98

IQR = 112

Lower boundary outlier(s) = 98 - 1.5(168)

Lower boundary outliers = -70

Thus there are no lower outliers.

Upper boundary outliers = 210 + 1.5(112)

Upper boundary outliers = 378

Therefore, 650 is an outlier.

<h3>4. The outlier(s) for these tires prices: $88, $135, $75, $135, $85, $168, $156, $192, $195, $210, $230, $245, $540</h3>

Arranging in increasing order:

75, 85, 88, 135, 135, 156, 168, 192, 195, 210, 230, 245, 540

Q1 = (88 + 135)/2

Q1 = 111.5

Q3 = (210 + 230)/2

Q3 = 220

IQR = 220 - 111.5

IQR = 108.5

Lower boundary outlier(s) = 111.5 - 1.5(108.5)

Lower boundary outliers = - 51.25

Thus, there are no lower outliers.

Upper boundary outliers = 220 + 1.5(111.5)

Upper boundary outliers = 387.25

Therefore, 540 is an outlier.

<h3>5. Quartiles for the tire pressures of car tires at an auto clinic: 23, 29,18, 24, 27, 24, 35, 32, 34</h3>

Arranging in increasing order:

18, 23, 24, 24, 27, 29, 32, 34, 35.

Q2 = 27

Q4 = 35

Q1 = (23 + 24) / 2

Q1 = 23.5

Q3 = (32 + 34) /2

Q3 = 33

Learn more about quartiles and outliers at: brainly.com/question/24805469

4 0

3 years ago

Solve x3 = 125 over 27. 25 over 9 ±25 over 9 5 over 3 ±5 over 3

gladu [14]

I hope this helps you

x.x.x=5.5.5

x=5

7 0

4 years ago

Does anyone know how to do this/have the answer? PLEASE

podryga [215]

15: You solve problems like this by finding the probability of each case, and then multiplying them all. For each of the last 4 question, she has probability 1/2 of guessing right. So, she guessed 4 consecutive questions with probability

$\left(\dfrac{1}{2}\right)^4=\dfrac{1}{16}$

16: Like before: you have pick a king with probability 4/52 = 1/13 (there are four kings - one for each suit, out of 52 cards in a standard deck), and you pick "I" from "INCREDIBLE" with probability 2/10 = 1/5 (there are two "I"s out of 10 letters). So, the probability of picking a king and then an "I" is

$\dfrac{1}{13}\cdot\dfrac{1}{5}=\dfrac{1}{65}$

17-20: The important bit of information here is that you replace the first ball. So, the first and second pick follow the exact same probability distribution, because they basically are two repetitions of the same experiment. So, for example, in ex. 17, the first ball is even with probability 15/30 = 1/2 (there are 15 even balls out of 30). Then, you have again probability 15/30 = 1/2 to pick an odd ball (there are also 15 odd balls out of 30). So, the probability of picking an even ball, replace it, and pick an odd ball is

$\dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{1}{4}$

Exercises 18 to 20 follow the same scheme: find out the probability of the two events and multiply them.

21-26: Not we DON'T replace the balls, so the second pick will suffer the effects of the first one. Let's dive into ex. 21 for example. For the first pick, we want a 2-digits number. There are 21 of such balls (all balls except balls 1 to 9), so we pick a 2-digits ball with probability 21/30 = 7/10. For the second pick, we want the balls number 4. But we have to assume that we already picked the first ball, and we picked a 2-digits ball. So, for the second pick, we're choosing from a bag with 29 balls, and there is only one ball labeled 4. So, we pick the ball number 4 with probability 1/29. We deduce that the two events happen one after the other with probability

$\dfrac{7}{10}\cdot\dfrac{1}{29}=\dfrac{7}{290}$

Exercises 22 to 26 are similar: you find out the probability of the first event, and then you consider the new environment (i.e. you keep track of the first pick) when it comes to the probability distribution for the second pick).

5 0

3 years ago

Which of the following angles are supplementary to 7? select all that apply.

max2010maxim [7]

Which of the following angles are supplementary to 7? select all that apply.

<u>A. 5</u>

B. 7

C. 6

<u>D. 8</u>

8 0

3 years ago

Help before i get my but beat

morpeh [17]

Answer:

top: 16

bottom: 9

side: 9

back 1: 16

back 2: 29

good luck

Step-by-step explanation:

6 0

3 years ago