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3 years ago

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After tossing the same coin 10 times, you are surprised to find that tails has come up 8 times. You therefore conclude that this

coin is not fair and that the probability of getting tails with this coin is 0.80.

1 answer:

Arisa [49]3 years ago

5 0

Whats the question you want us to solve

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How to write the number 1.39-1.14 in word form

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0.25 which is a quarter.

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4 years ago

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Find the equation in stope intercept form of the line that passes through the points (-2,4) and (6,2),

Juli2301 [7.4K]

Answer:

y = -x/4 + 7/2

Step-by-step explanation:

Find the slope

( -2 , 4) ( 6 , 2)

m = (y2 - y1)/(x2 - x1)

x1 = -2

y1 = 4

x2 = 6

y2 = 2

m = ( 2 - 4)/( 6 -(-2)

m = ( 2 -4)/(6 +2)

= -2/8

= -1/4

Using point slope form equation

y - y1 = m( x - x1)

Using the second point

( 6 , 2)

x1= 6

y1=2

m = -1/4

y - 2 = -1/4(x - 6)

y - 2 = -(x - 6)/4

Open the bracket with -

y - 2 = (-x + 6)/4

y = ( -x + 6)/4 + 2

LCM = 4

y =( -x + 6 + 8)/4

y = (-x + 14)/4

Rearrange using slope intercept form

y = mx + c.

y = -x/4 + 14/4

We can break 14/4 by dividing by 2

y = -x/4 + 7/2

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4 years ago

What is the mean (average) of 2,4,5,6,8

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Answer:

5

Step-by-step explanation:

In order to figure out average, we need to add up all the numbers. So adding 2, 4, 5, 6, and 8 gives us 25. We now divide that total by the total number of terms we used to reach 25. So since we had 5 numbers, we divide 25 by 5 to get an average of 5.

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4 years ago

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9. On a game show, there are 16 questions: 8 easy, 5 medium-hard, and 3 hard. If contestants are given questions

just olya [345]

Answer:

$\frac{7}{30}\approx0.23$

Step-by-step explanation:

Given:

Number of questions (N) = 16

Number of easy questions (E) = 8

Number of medium-hard questions (M) = 5

Number of hard questions (H) = 3

Now, the probability of getting the first question as easy question is given as:

$P(E1)=\frac{\textrm{Number of easy questions}}{\textrm{Total questions}}\\\\P(E1)=\frac{E}{N}=\frac{8}{16}=0.5$

Now, probability of getting the second question as easy question is given as:

$P(E2)=\frac{\textrm{Number of easy questions left}}{\textrm{Total questions left}}\\\\P(E2)=\frac{E-1}{N-1}=\frac{7}{15}$

Now, probability that the first two contestants will get easy questions is given by the product of $P(E1)\ and\ P(E2)$ . So,

$P(2\ easy\ questions)=P(E1)\times P(E2)\\\\P(2\ easy\ questions)=\frac{1}{2}\times \frac{7}{15}\\\\P(2\ easy\ questions)=\frac{7}{30}\approx 0.23$

Therefore, the probability that the first two contestants will get easy questions is $\frac{7}{30}\approx0.23$

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3 years ago

Five companies (A, B, C, D, and E) that make elec- trical relays compete each year to be the sole sup- plier of relays to a majo

NNADVOKAT [17]

Answer:

a

$P(a | e') = 0.22$

$P(b | e') = 0.28$

$P(c | e') = 0.33$

b

$P(a | e' , d' , b') = 0.57$

Step-by-step explanation:

From the question we are told that

The probabilities are

Supplier chosen A B C

Probability P(a) = 0.20 P(b) = 0.25 P(c) = 0.15

D E

P(d) = 0.30 P(e) = 0.10

Generally the new probability of companies A being chosen as the sole supplier this year given that supplier E goes out of business is mathematically represented as below according to Bayes theorem

$P(a | e') = \frac{P (a \ and \ e')}{P(e')}$

$P(a | e') = \frac{P (a)}{P(e')}$

$P(a | e') = \frac{P (a)}{1- P(e)}$

=> $P(a | e') = \frac{ 0.20}{1- 0.10}$

=> $P(a | e') = 0.22$

Generally the new probability of companies B being chosen as the sole supplier this year given that supplier E goes out of business is mathematically represented as below according to Bayes theorem

$P(b | e') = \frac{P (b \ and \ e')}{P(e')}$

$P(b | e') = \frac{P (b)}{P(e')}$

$P(b | e') = \frac{P (b)}{1- P(e)}$

=> $P(b | e') = \frac{ 0.25}{1- 0.10}$

=> $P(b | e') = 0.28$

Generally the new probability of companies C being chosen as the sole supplier this year given that supplier E goes out of business is mathematically represented as below according to Bayes theorem

$P(c | e') = \frac{P (c \ and \ e')}{P(e')}$

$P(c | e') = \frac{P (c)}{P(e')}$

$P(c | e') = \frac{P (c)}{1- P(e)}$

=> $P(c | e') = \frac{ 0.15}{1- 0.10}$

=> $P(c | e') = 0.17$

Generally the new probability of companies D being chosen as the sole supplier this year given that supplier E goes out of business is mathematically represented as below according to Bayes theorem

$P(d | e') = \frac{P (d \ and \ e')}{P(e')}$

$P(d | e') = \frac{P (d)}{P(e')}$

$P(d | e') = \frac{P (d)}{1- P(e)}$

=> $P(d | e') = \frac{ 0.30}{1- 0.10}$

=> $P(c | e') = 0.33$

Generally the probability that B, D , E are not chosen this year is mathematically represented as

$P(N) = 1 - [P(e) +P(b) + P(d) ]$

=> $P(N) = 1 - [0.10 +0.25 +0.30 ]$

=> $P(N) = 0.35$

Generally the probability that A is chosen given that E , D , B are rejected this year is mathematically represented as

$P(a | e' , d' , b') = \frac{P(a)}{P(N)}$

=> $P(a | e' , d' , b') = \frac{0.20 }{0.35 }$

=> $P(a | e' , d' , b') = 0.57$

5 0

3 years ago