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MakcuM [25]
3 years ago
6

Jason and Maria are playing a board game in which three dice are tossed to determine a player's move. Find each probability for:

P (two 5s) , P (three 5s) , and P (one 5 or two 5s).
please show work :)
Mathematics
1 answer:
fredd [130]3 years ago
6 0
Tossing a die will have 6 possible outcomes. Those are having sides that are number 1 to 6. The sample space of tossing 3 dice is equal to 6³ which is equal to 216. Now for the calculation of probabilities,

P(two 5s) = (1 x 1 x 5)/216
As we have to have the 5 in the die for two times, then for the 1 time, we can have all other numbers except 5. The answer is 5/216. 

P(three 5s) = (1 x 1 x 1)/216   = 1/216

P(one 5 or two 5s)  = (1 x 5 x 5)/216 + (1 x 1 x 5)/216 = 5/36
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the number is 57,733 contains two sets of digits in which one digit is ten times as great as the other. what are the values of t
Afina-wow [57]
In 57,733 the values are:

5 is for tens
The first 7 is for ones
The second 7 is for tenths
The first 3 is for hundredths
The second 3 is for thousandths

Two set of digit in which one digit is ten times as great as the other are 3 and 3, as well as 7 and 7

It's because 0,03 is ten times as much as 0,003 and 7 is ten times as much as 0,7.
8 0
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liam deposits $2400 into an account that earns 8.3% interest compounded quarterly. how much money will liam have in 5 years.
Harlamova29_29 [7]

Answer:

Total = 2,400 * (1 + (.083/4))^4*5

Total = 2,400 * (1.02075)^20

Total = 2,400 * 1.5079528829

Total = 3,619.09

Step-by-step explanation:

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3 years ago
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Evaluate the function f(x)=tan^-1(x) when x=0. Give your answer in radians.
vichka [17]
The answer is equal to 0.
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Which of the following values for x and y satisfy the following system of the equations?
weqwewe [10]

Answer:

The answer to your question is letter C

Step-by-step explanation:

Equations

                             x  +  4y  =  10

                           5x + 10y  = 20

Process

1.- Substitute all the options in both equations and evaluate them

a) x = 3, y = 2              This option is incorrect

           (3) +  4(2)  =  10             3 + 8 = 10         11 ≠ 10

         5(3) + 10(2)  = 20           15 + 20 = 20      35 ≠ 20    

b) x = 2, y = -3       This option is incorrect

  (2)  +  4(-3)  =  10                  2 -12 = 10            -10 ≠ 10

  (2) + 10(-3)  = 20                   2 - 30 = 20        -28 ≠ 20

c)  x = -2, y = 3    This is the right answer

  (-2)  +  4(3)  =  10                -2 + 12 = 10            10 = 10

 5(-2) + 10(3)  = 20               -10 + 30 = 20          20 = 20

d) x = 3, y = -2               This option is incorrect

  (3)  +  4(-2) =  10                 3 - 6 = 10                 - 3 ≠ 10

 5(3) + 10(-2)  = 20              15 - 20 = 20              -10 ≠ 10

6 0
3 years ago
In a population of similar households, suppose the weekly supermarket expense for a typical household is normally distributed wi
Rina8888 [55]

Answer:

P(Y ≥ 15) = 0.763

Step-by-step explanation:

Given that:

Mean =135

standard deviation = 12

sample size n  = 50

sample mean \overline x = 140

Suppose X is the random variable that follows a normal distribution which represents the weekly supermarket expenses

Then,

X \sim N ( \mu \sigma)

The probability that X is greater than 140 is :

P(X>140) = 1 - P(X ≤ 140)

P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{140-135}{12})

P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{5}{12})

P(X>140) = 1 - P( Z\leq0.42)

From z tables,

P(X>140) = 1 - 0.6628

P(X>140) = 0.3372

Similarly, let consider Y to be the variable that follows a binomial distribution of the no of household whose expense is greater than $140

Then;

Y \sim Binomial (np)

Y \sim Binomial (50,0.3372)

∴

P(Y ≥ 15) = 1- P(Y< 15)

P(Y ≥ 15) = 1 - ( P(Y=0) + P(Y=1) + P(Y=2) + ... + P(Y=14) )

P(Y \geq 15) = 1 - \begin {pmatrix} ^{50}_0 \end {pmatrix} (0.3372)^0 (1-0,3372)^{50} + \begin {pmatrix} ^{50}_1 \end {pmatrix} (0.3372)^1 (1-0,3372)^{49}  + \begin {pmatrix} ^{50}_2 \end {pmatrix} (0.3372)^2 (1-0,3372)^{48} +...  + \begin {pmatrix} ^{50}_{50{ \end {pmatrix} (0.3372)^{50} (1-0,3372)^{0}

P(Y ≥ 15) = 0.763

7 0
3 years ago
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