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You roll a number cube twice. Find the probability of the events. Rolling an even number of 5.

lys-0071 [83]

Answer:

An even number and a 5. In the first roll you can get 2, 4 or 6, so you have 3 out of 6 options, the probability is 3/6 = 1/2, in the second roll you need you get a 5, the probability is 1/6.

I really hope this helps please give brainliest!!!!

5 0

3 years ago

Here is a simple probability model for multiple-choice tests. Suppose that each student has probability p of correctly answering

Alla [95]

Answer:

a) The probability that Jodi scores 78% or lower on a 100-question test is 4%.

b) The probability that Jodi scores 78% or lower on a 250-question test is 0.023%.

Step-by-step explanation:

a) To approximate this distribution we have to calculate the mean and the standard distribution.

The mean is the proportion p=0.85.

The standard deviation can be calculates as:

$\sigma=\sqrt{\frac{p(1-p)}{n} }= \sqrt{\frac{0.85*(1-0.85)}{100} }=0.04$

To calculate the probability that Jodi scores 78% or less on a 100-question test, we first calculate the z-value:

$z=\frac{p-p_0}{\sigma} =\frac{0.78-0.85}{0.04} =-1.75$

The probability for this value of z is

$P(x$

The probability that Jodi scores 78% or lower on a 100-question test is 4%.

b) In this case, the number of questions is 250, so the standard deviation needs to be calculated again:

$\sigma=\sqrt{\frac{p(1-p)}{n} }= \sqrt{\frac{0.85*(1-0.85)}{250} }=0.02$

To calculate the probability that Jodi scores 78% or less on a 250-question test, we first calculate the z-value:

$z=\frac{p-p_0}{\sigma} =\frac{0.78-0.85}{0.02} =-3.5$

The probability for this value of z is

$P(x$

The probability that Jodi scores 78% or lower on a 250-question test is 0.023%.

6 0

3 years ago

Estimate 26% of 84. 26% ≈ 84 ≈ 80 So, 26% of 84 ≈ .

Juli2301 [7.4K]

Answer:

21

Step-by-step explanation:

8 0

2 years ago

If (x, y) = (3, 4) is a solution of the simultaneous equations.

Vinvika [58]

The first thing you want to do is plug in x and y into both equations:

a(3) + b(4) = 4

b(3) + a(4) = 8

rearrange to line up a’s and b’s

3a + 4b = 4

4a + 3b = 8

now you want to choose a or b and multiply each equation by a number to make them have the same amount of a’s or b’s.

4(3a + 4b = 4) = 12a + 16b = 16

3(4a + 3b = 8) = 12a + 9b = 24

Now we subtract the bottom equation from the top and solve for b:

12a + 16b - (12a + 9b) = 16 - 24

7b = -8

b = -8/7

Now we plug back in for b to one of the original equations:

3a + 4(-8/7) = 4

3a + (-32/7) = 4

3a - (32/7) = 4

3a = 4 + (32/7)

3a = (28/7) + (32/7)

3a = 60/7

a = (60/7)/3 = 20/7.

Finally, plug a and b in together to double check using the second equation.

4a + 3b = 8

4(20/7) + 3(-8/7) = ?

(80/7) - (24/7) = ?

56/7 = 8.

4 0

3 years ago

7x + y = 7 y = [ ? ]x+ []

otez555 [7]

Y=-7x+7

hope this helped!

6 0

3 years ago