we conclude that at 5:00 p.m. there are 8 more inches of snow than at 8:00 a.m.
<h3>How many more inches of snow were on the ground at 5:00 p.m. than at 8:00 a.m.?</h3>
We know that at 8:00 a.m. there were t inches of snow in the ground.
At 5:00 p.m. there were 3t inches of snow in the ground.
Then the difference between the heights of the snow is:
3t- t = 2t
And we know that at 5:00 p.m. there were 12 inches of snow then we can solve the linear equation for t:
3t = 12in
t = (12in)/3 = 4 in
Replacing that in the difference of heights:
2t = 2*4in = 8in
From this, we conclude that at 5:00 p.m. there are 8 more inches of snow than at 8:00 a.m.
If you want to learn more about linear equations:
brainly.com/question/1884491
#SPJ1
Answer:
the 6x element as there is no solution in integers.
Step-by-step explanation:
Answer:
C. n=23; p^=0.5
Step-by-step explanation:
Normal distribution is symmetrical about the mean.
So, p should be close to ½
Answer:
c) Is not a property (hence (d) is not either)
Step-by-step explanation:
Remember that the chi square distribution with k degrees of freedom has this formula

Where N₁ , N₂m ....
are independent random variables with standard normal distribution. Since it is a sum of squares, then the chi square distribution cant take negative values, thus (c) is not true as property. Therefore, (d) cant be true either.
Since the chi square is a sum of squares of a symmetrical random variable, it is skewed to the right (values with big absolute value, either positive or negative, will represent a big weight for the graph that is not compensated with values near 0). This shows that (a) is true
The more degrees of freedom the chi square has, the less skewed to the right it is, up to the point of being almost symmetrical for high values of k. In fact, the Central Limit Theorem states that a chi sqare with n degrees of freedom, with n big, will have a distribution approximate to a Normal distribution, therefore, it is not very skewed for high values of n. As a conclusion, the shape of the distribution changes when the degrees of freedom increase, because the distribution is more symmetrical the higher the degrees of freedom are. Thus, (b) is true.