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9966 [12]
3 years ago
7

You select a card at random from the cards that make up the word “replacement”. Without replacing the card, you choose a second

card. Find the probability of choosing a consonant and then an “e”. There is 1 letter for each card.
Mathematics
1 answer:
natta225 [31]3 years ago
6 0
To find the total probability, we first need to solve for the probability of the individual events.

Event 1:
P(consonant)=7/11
We know this to be true because out of the 11 total possibilities for the event, 7 of them are consonants.
R E P L A C E M E N T 


Event 2
P(e)=3/10
We know this to be true because out of the 10 total possibilities for the event (since we didn't replace the first card we withdrew), 3 of them are the letter 'e'.
R E P L A C E M E N T (-1 to account for the first card we withdrew)

The possibility of both events...

P(consonant then 'e')=7/11*3/10=21/110

Answer: P=21/110
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The mean shoe size of the students in a math class is 7. 5. Most of the shoe sizes fall within 1 standard deviation, or between
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Answer: 1.5

Step-by-step explanation:Let the standard deviation of the shoe size be represented by ' ' and the mean shoe size be represented by ' = 7.5'. Also, the sizes of the shoe be represented by ' '. So, . Hence, the standard deviation of the shoe size data for the math class is 1.5.

4 0
2 years ago
I think of a number and on adding 13 to it ,I get 40.Write the equation for it ?​
Alekssandra [29.7K]

Answer:

x+13=40 and the answer to that is 27

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What does X equal x+10
Ilya [14]
It’s already simplify
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3 years ago
How do you figure out 3h = 120
cestrela7 [59]

Answer:

h=40

Step-by-step explanation:

3h = 120

Divide 3 on both sides of the equation.

\frac{3h}{3}=\frac{120}{3}

h=40

5 0
3 years ago
Read 2 more answers
medical tests. Task Compute the requested probabilities using the contingency table. A group of 7500 individuals take part in a
uysha [10]

Probabilities are used to determine the chances of an event

  • The probability that a person is sick is: 0.008
  • The probability that a test is positive, given that the person is sick is 0.9833
  • The probability that a test is negative, given that the person is not sick is: 0.9899
  • The probability that a person is sick, given that the test is positive is: 0.4403
  • The probability that a person is not sick, given that the test is negative is: 0.9998
  • A 99% accurate test is a correct test

<u />

<u>(a) Probability that a person is sick</u>

From the table, we have:

\mathbf{Sick = 59+1 = 60}

So, the probability that a person is sick is:

\mathbf{Pr = \frac{Sick}{Total}}

This gives

\mathbf{Pr = \frac{60}{7500}}

\mathbf{Pr = 0.008}

The probability that a person is sick is: 0.008

<u>(b) Probability that a test is positive, given that the person is sick</u>

From the table, we have:

\mathbf{Positive\ and\ Sick=59}

So, the probability that a test is positive, given that the person is sick is:

\mathbf{Pr = \frac{Positive\ and\ Sick}{Sick}}

This gives

\mathbf{Pr = \frac{59}{60}}

\mathbf{Pr = 0.9833}

The probability that a test is positive, given that the person is sick is 0.9833

<u>(c) Probability that a test is negative, given that the person is not sick</u>

From the table, we have:

\mathbf{Negative\ and\ Not\ Sick=7365}

\mathbf{Not\ Sick = 75 + 7365 = 7440}

So, the probability that a test is negative, given that the person is not sick is:

\mathbf{Pr = \frac{Negative\ and\ Not\ Sick}{Not\ Sick}}

This gives

\mathbf{Pr = \frac{7365}{7440}}

\mathbf{Pr = 0.9899}

The probability that a test is negative, given that the person is not sick is: 0.9899

<u>(d) Probability that a person is sick, given that the test is positive</u>

From the table, we have:

\mathbf{Positive\ and\ Sick=59}

\mathbf{Positive=59 + 75 = 134}

So, the probability that a person is sick, given that the test is positive is:

\mathbf{Pr = \frac{Positive\ and\ Sick}{Positive}}

This gives

\mathbf{Pr = \frac{59}{134}}

\mathbf{Pr = 0.4403}

The probability that a person is sick, given that the test is positive is: 0.4403

<u>(e) Probability that a person is not sick, given that the test is negative</u>

From the table, we have:

\mathbf{Negative\ and\ Not\ Sick=7365}

\mathbf{Negative = 1+ 7365 = 7366}

So, the probability that a person is not sick, given that the test is negative is:

\mathbf{Pr = \frac{Negative\ and\ Not\ Sick}{Negative}}

This gives

\mathbf{Pr = \frac{7365}{7366}}

\mathbf{Pr = 0.9998}

The probability that a person is not sick, given that the test is negative is: 0.9998

<u>(f) When a test is 99% accurate</u>

The accuracy of test is the measure of its sensitivity, prevalence and specificity.

So, when a test is said to be 99% accurate, it means that the test is correct, and the result is usable; irrespective of whether the result is positive or negative.

Read more about probabilities at:

brainly.com/question/11234923

4 0
3 years ago
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