(71+x)/2=78
71+x=156
x=85
Th student needs to score a min of 85
For this case, we have that by definition, a budget refers to:
<em>An advance calculation of the cost of a work or a service.
</em>
For example:
"I asked the mechanic for the repair budget for my vehicle".
As another definition we have:
<em>A set of expenses and income expected for a certain period of time. In the same way, it can be defined as a plan of operations and resources of a company.</em>
<em />
Log₂(6x) - log₂(√x) = 2
log₂[6x/√x] = 2, let's rewrite it in terms of exponent:
(remember : ㏒₂(x) = a (in exponential form) → x = 2ᵃ
6x/√x = 2²
6x/√x = 4, square both sides:
36x²/x = 16 .Simplify:
36x = 16 and x = 16/36 . So x = 4/9
This Is One Answer, There Are More Then One Answer.
100 / 5 = 20 + 27 = 47 / 2 = 23.5
100 = my number
5 = divide by
100 / 5 = 20
27 = add
20 = results of last answer
Then 20 + 27 = 47
47 = results of last answer
2 = half of
Then 47 / 2 = 23.5
Answer:
If we compare the p value and using the significance level given
we have
so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% or 1% of significance we fail to reject the null hypothesis.
Step-by-step explanation:
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level is not provided but we can assume it as
. First we need to calculate the degrees of freedom like this:

The next step would be calculate the p value for this test. Since is a bilateral test or two tailed test, the p value would be:
If we compare the p value and using the significance level given
we have
so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% or 1% of significance we fail to reject the null hypothesis.