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Novay_Z [31]
2 years ago
6

An automobile manufacturer is concerned about a possible recall of its best selling four-door sedan. If there was a recall there

is a probability that a defect is in the brake system, in the transmission, in the fuel system and in some other area. Part 1 What is the probability that the defect is in the brakes or the transmission system if the probability of defect in both the systems simultaneously is
Mathematics
1 answer:
goldfiish [28.3K]2 years ago
7 0

Complete question is;

An automobile manufacturer is concerned about a possible recall of its best - selling four-door sedan. If there were a recall, there is a probability of 0.25 of a defect in the brake system, 0.18 of a defect in the transmission, 0.17 of a defect in the fuel system, and 0.40 of a defect in some other area.

(a) What is the probability that the defect is the brakes or the fueling system if the probability of defects in both systems simultaneously is 0.15?

(b) What is the probability that there are no defects in either the brakes or the fueling system?

Answer:

A) 0.27

B) 0.73

Step-by-step explanation:

Let B denote that there is a defect in the break system

Let T demote that there is a defect in transmission

Let F denote that there is a defect in the fuel system.

Let P denote that there is a defect in some other area.

Now, we are given;

P(B) = 0.25

P(T) = 0.18

P(F) = 0.17

P(O) = 0.40

A) We are given that probability of defects in both brakes and the fueling system simultaneously is 0.15.

Thus, it means; P(B ⋂ F) = 0.15

Now, we want to find the probability that the defect is the brakes or the fueling system. This is expressed as;

P(B ⋃ F) = P(B) + P(F) - P(B ⋂ F)

Plugging in the relevant values to give;

P(B ⋃ F) = 0.25 + 0.17 - 0.15

P(B ⋃ F) = 0.27

B) We want to find the probability that there are no defects in either the brakes or the fueling system.

This is expressed as;

P(B' ⋃ F') = 1 - P(B ⋃ F)

Plugging in relevant value;

P(B' ⋃ F') = 1 - 0.27

P(B' ⋃ F') = 0.73

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anyanavicka [17]

Answer:

(2, 5)

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Equality Properties

<u>Algebra I</u>

  • Solving systems of equations using substitution/elimination

Step-by-step explanation:

<u>Step 1: Define Systems</u>

y - 3x = 1

2y - x = 12

<u>Step 2: Rewrite Systems</u>

y - 3x = 1

  1. Add 3x on both sides:                    y = 3x + 1

<u>Step 3: Redefine Systems</u>

y = 3x + 1

2y - x = 12

<u>Step 4: Solve for </u><em><u>x</u></em>

<em>Substitution</em>

  1. Substitute in <em>y</em>:                    2(3x + 1) - x = 12
  2. Distribute 2:                         6x + 2 - x = 12
  3. Combine like terms:           5x + 2 = 12
  4. Isolate <em>x</em> term:                     5x = 10
  5. Isolate <em>x</em>:                              x = 2

<u>Step 5: Solve for </u><em><u>y</u></em>

  1. Define equation:                    2y - x = 12
  2. Substitute in <em>x</em>:                       2y - 2 = 12
  3. Isolate <em>y </em>term:                        2y = 10
  4. Isolate <em>y</em>:                                 y = 5
4 0
2 years ago
Two stores sell the same computer for the same original price. Store A advertises that the computer is on sale for 25% off the o
arlik [135]
N= P-180. so the computer would originally be 720 dollars. reduced by 25%, or 180, the new price is 540.
5 0
3 years ago
Read 2 more answers
A statistics student wants to compare his final exam score to his friend's final exam score from last year; however, the two exa
yarga [219]

Answer:

z= \frac{85-70}{10}=1.5

z= \frac{45-35}{5}=2

So then the correct answer for this case is:

B) Our student, Z= 1.50; his friend, Z=2.00.

Step-by-step explanation:

Assuming this complete question:

A statistics student wants to compare his final exam score to his friend's final exam score from last year; however, the two exams were scored on different scales. Remembering what he learned about the advantages of Z scores, he asks his friend for the mean and standard deviation of her class on the exam, as well as her final exam score. Here is the information:

Our student: Final exam score = 85; Class: M = 70; SD = 10.

His friend: Final exam score = 45; Class: M = 35; SD = 5.

The Z score for the student and his friend are:

A) Our student, Z= -1.07; his friend, Z= -1.14.

B) Our student, Z= 1.50; his friend, Z=2.00.

C) Our student, Z= 1.07; his friend, Z= -1.14.

D) Our student, Z= 1.07; his friend, Z= 1.50

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution

Let X the random variable that represent the scores for our student, and for this case we know that:

E(X)= \mu =70, SD_X=\sigma=10  

The z score is given by:

z=\frac{x-\mu}{\sigma}

If we use this we got:

z= \frac{85-70}{10}=1.5

Let Y the random variable that represent the scores for his friend, and for this case we know that:

E(Y)= \mu =35, SD_Y=\sigma=5  

The z score is given by:

z=\frac{y-\mu}{\sigma}

If we use this we got:

z= \frac{45-35}{5}=2

So then the correct answer for this case is:

B) Our student, Z= 1.50; his friend, Z=2.00.

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The value of the computer is given by the expression 2000 + (200t) where t is the time in years
OLEGan [10]
If it was one year it would be 2200 and you just keep adding 200 for each year.
4 0
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What is the absolute mean deviation of this data?<br> 1, 2, 3, 3, 4,5, 6, 8
Harlamova29_29 [7]

Answer:

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Step-by-step explanation:

First add all the numbers

1+2+3+4+5+6+8= 29

Finally you divide the total by counting the numbers which you have add up

29÷7

= 4/1/7 ( 4 and 1 over 7)

Mark me as the brainiest answer

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2 years ago
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