The table shows the results of rolling a number cube, with sides labeled 1 through 6, several times.

Mathematics

1 answer:

mr_godi [17]2 years ago

5 0

$|\Omega|=10+6+4+8+6+6=40\\ |A|=10+6=16\\\\ P(A)=\dfrac{16}{40}=\dfrac{2}{5}=40\%$

You might be interested in

2x18-5+15%5=32<br><br> Insert parentheses to make each equal true

Colt1911 [192]

Answer:

( 2×8) - [ ( 5 + 15 ) ÷ 5] = 32

Step-by-step explanation:

According to the scenario, computation of the given data are as follows,

Given equation = 2x18-5+15/5=32

First we multiply first two letters,

then ( 2 × 18) = 36

Now, we add 5 and 15 then divide it by 5,

So, ( 5 + 15 ) ÷ 5 = 20 ÷ 5 = 4

Now we subtract 4 from 36,

Then 36 - 4 = 32

Hence the correct parentheses = ( 2×8) - [ ( 5 + 15 ) ÷ 5] = 32

3 0

2 years ago

The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78

____ [38]

Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

$m-t_{a/2,n-1}\frac{s}{\sqrt{n} }$ ≤ μ ≤ $m+t_{a/2,n-1}\frac{s}{\sqrt{n} }$

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and $t_{a/2,n-1}$ is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and $t_{0.05,9}$ by 1.8331, we get that the 90% confidence interval for the mean speed is: