A fair die is rolled 3 times what is the probability that is comes up 1 at least Once

Mathematics

1 answer:

stiv31 [10]2 years ago

4 0

I think it would be 1/2 because the probability out of 1 try is 1/6 and 1/6 times 3 is 1/2

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Solve the equation using elimination. Is this a consistent, inconsistent, or dependent? What is the solution

Basile [38]

9514 1404 393

Answer:

(x, y) = (-1, -3)

Step-by-step explanation:

The equations are "consistent" and "not dependent." This will be the case whenever the ratios of x- and y-coefficients are different.

We can solve this by "elimination" by multiplying the first equation by -4 and adding the result of the second equation being multiplied by 7.

-4(2x -7y) +7(7x -4y) = -4(19) +7(5)

-8x +28y +49x -28y = -76 +35 . . . . eliminate parentheses

41x = -41 . . . . . simplify

x = -1 . . . . . . . divide by 41

Using the second equation, we find y to be ...

(7x -5)/4 = y = (7(-1) -5)/4 = -12/4 = -3

So, the solution is (x, y) = (-1, -3).

7 0

2 years ago

A survey was conducted that asked 1003 people how many books they had read in the past year. Results indicated that x overbar eq

Sergio [31]

Answer:

The 99% confidence interval would be given (11.448;14.152).

Step-by-step explanation:

1) Important concepts and notation

A confidence interval for a mean "gives us a range of plausible values for the population mean. If a confidence interval does not include a particular value, we can say that it is not likely that the particular value is the true population mean"

$s=16.6$ represent the sample deviation

$\bar X=12.8$ represent the sample mean

n =1003 is the sample size selected

Confidence =99% or 0.99

$\alpha=1-0.99=0.01$ represent the significance level.

2) Solution to the problem

The confidence interval for the mean would be given by this formula

$\bar X \pm z_{\alpha/2} \frac{s}{\sqrt{n}}$

We can use a z quantile instead of t since the sample size is large enough.

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.