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vagabundo [1.1K]
3 years ago
15

4 Fernanda is rolling dice. What is the probability of rolling snake eyes "roll both ones?

Mathematics
1 answer:
tresset_1 [31]3 years ago
8 0

Answer:

1/36

Step-by-step explanation:

The odds of rolling one on a single die are 1/6. It gives (1/6)(1/6) = 1/36 to multiply this by the probability of rolling one on a second die.

If you look at all possible rolling two dice combinations, there are 36. Of those 36, only one is the "snake eyes." Therefore, the probability of "snake eyes" rolling is 1/36.

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A school requires 2 computers for every 5 students. Write a proportion that gives the number c of computers needed for 145 stude
guajiro [1.7K]
The proportion would go as follows:

2   C
5 = 145      Its hard to type that but 2 would go over the 5 and an equals sign would be in the middle, then C over 45. To solve then take 145 times 2 and divide it by 5.
Hope this helps. 
5 0
3 years ago
The distance from Jason’s house to school is 0.5 km what is the distance in meters
valkas [14]

Answer:

500 meters

Step-by-step explanation:

1km=1000meters so 0.5km=500m

7 0
3 years ago
Solve the equation:<br> [3 2<br> 5 5]<br><br> [x1<br> x2]+<br> [1<br> 2]=<br> [2<br> -3]
dsp73

9514 1404 393

Answer:

  (x1, x2) = (3, -4)

Step-by-step explanation:

As with any 2-step linear equation, subtract the constant, then multiply by the inverse of the coefficient of the variable.

  \left[\begin{array}{cc}3&2\\5&5\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]+\left[\begin{array}{c}1\\2\end{array}\right]=\left[\begin{array}{c}2\\-3\end{array}\right]\\\\\left[\begin{array}{cc}3&2\\5&5\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{c}1\\-5\end{array}\right]\\\\\left[\begin{array}{c}x\\y\end{array}\right]=\dfrac{1}{5}\left[\begin{array}{cc}5&-2\\-5&3\end{array}\right]\left[\begin{array}{c}1\\-5\end{array}\right]

Performing the multiplication of the matrix by the vector gives the solution.

  x = ((5)(1) +(-2)(-5))/5 = 15/5 = 3

  y = ((-5)(1) +(3)(-5))/5 = -20/5 = -4

Using your variables, x1, x2, the solution is ...

  (x1, x2) = (3, -4)

7 0
3 years ago
Write in standard form
ValentinkaMS [17]
34.06 is the the correct way of saying it.
6 0
3 years ago
A cable car starts off with n riders. The times between successive stops of the car are independent exponential random variables
nikitadnepr [17]

Answer:

The distribution is \frac{\lambda^{n}e^{- \lambda t}t^{n - 1}}{(n - 1)!}

Solution:

As per the question:

Total no. of riders = n

Now, suppose the T_{i} is the time between the departure of the rider i - 1 and i from the cable car.

where

T_{i} = independent exponential random variable whose rate is \lambda

The general form is given by:

T_{i} = \lambda e^{- lambda}

(a) Now, the time distribution of the last rider is given as the sum total of the time of each rider:

S_{n} = T_{1} + T_{2} + ........ + T_{n}

S_{n} = \sum_{i}^{n} T_{n}

Now, the sum of the exponential random variable with \lambda with rate \lambda is given by:

S_{n} = f(t:n, \lamda) = \frac{\lambda^{n}e^{- \lambda t}t^{n - 1}}{(n - 1)!}

5 0
3 years ago
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