Given that Allison’s dog is 6 pounds more than 5 times the weight of Gail’s dog, it is important to know the following:
1. The word "more" indicates an Addition.
2. The word "times" indicates Multiplication.
Knowing this, you can write the following expression to represent "5 times the weight of Gail’s dog", where "g" is the weight of Gail’s dog:
Therefore, you can determine that the weight of Allison's dog (in pounds) is the sum of 6 pounds and 5 times the weight of Gail's dog (in pounds). Then, you can represent this with this algebraic expression:
Hence, the answer is:
The answer to this question will depend on the function f itself. Basically you will find the height in meters above the ground of the bird when it jumped when the time t=0s. This is substsitute every t in the function for a value of zero and that way you will get the bird's height at the time it jumped. If you were given a graph for this function, you can find the y-intercept of the graph and that will be the answer as well. The question could be written like this:
A baby bird jumps from a tree branch and flutters to the ground. The function "
" models the bird's height (in meters) above the ground as a function of time (in seconds) after jumping. What is bird's height above the ground when it jumped.
Answer:
25m
Step-by-step explanation:
Once your function is given, you can substitute t=0 since 0s is the time measured at the moment the bird jumped. So our function will be:
So the height of the bird above the ground when it jumped is 25m in this particular function.
Answer:
1/52
Step-by-step explanation:
Answer:
+/- 1.9983
Step-by-step explanation:
The critical values for a confidence interval for the population mean, with population standard deviation not known, are obtained from the student's t distribution.
The sample size is given as 64. The degrees of freedom will thus be;
sample size - 1 = 64 - 1 = 63
The confidence level is given as 95%. The level of significance will thus be;
100 - 95 = 5%.
The area in the right-tail will thus be 2.5%. Therefore, we looking for a t-value such that the area to its right is 2.5% and the degrees of freedom being 63. From the t-tables we have;
+/- 1.9983
