Answer:
15/4 please let me know if this helped you :)
Step-by-step explanation:
0.75*5=3.75 or 15/4
all of your answers are less than 3/4ths of a mile...she is increasing the amount she runs not decreasing.
Answer:
Range is {y | y ≥ –11}
Step-by-step explanation:
This is quadratic equation.
<em>A quadratic equation's range can be found if we find the vertex.</em>
For quadratic equations that have a positive number in front of
, it is upward opening and thus <u>all the numbers greater than or equal to the minimum value of vertex is the range.</u>
The formula for vertex of a parabola is:
Vertex = 
Where,
is the coefficient of 
is the coefficient of 
From our equation given,
and 
Now,
coordinate of vertex is 
coordinate of the vertex IS THE MINIMUM VALUE that we want. We get this by plugging in the
value [
] into the equation. So we have:

Hence, the range would be all numbers greater than or equal to
Third answer choice is the right one.
Answer:
![12-[20-2(6^2\div3\times2^2)]=88](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D%3D88)
Step-by-step explanation:
So we have the expression:
![12-[20-2(6^2\div3\times2^2)]](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D)
Recall the order of operations or PEMDAS:
P: Operations within parentheses must be done first. On a side note, do parentheses before brackets.
E: Within the parentheses, if exponents are present, do them before all other operations.
M/D: Multiplication and division next, whichever comes first.
A/S: Addition and subtraction next, whichever comes first.
(Note: This is how the order of operations is traditionally taught and how it was to me. If this is different for you, I do apologize. However, the answer should be the same.)
Thus, we should do the operations inside the parentheses first. Therefore:
![12-[20-2(6^2\div3\times2^2)]](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D)
The parentheses is:

Square the 6 and the 4:

Do the operations from left to right. 36 divided by 3 is 12. 12 times 4 is 48:

Therefore, the original equation is now:
![12-[20-2(6^2\div3\times2^2)]\\=12- [20-2(48)]](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D%5C%5C%3D12-%20%5B20-2%2848%29%5D)
Multiply with the brackets:
![=12-[20-96]](https://tex.z-dn.net/?f=%3D12-%5B20-96%5D)
Subtract with the brackets:
![=12-[-76]](https://tex.z-dn.net/?f=%3D12-%5B-76%5D)
Two negatives make a positive. Add:

Therefore:
![12-[20-2(6^2\div3\times2^2)]=88](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D%3D88)
The answers to the questions are:
1. The confidence level is 99 percent.
2. We have to conclude that there is no sufficient evidence available to support this claim because the Confidence interval contains 75 sec.
<h3>How to solve for the confidence level</h3>
1. The confidence level here should be
1- 0.01 = 0.99
= 99 percent
Given that, 99% confidence interval for population mean (μ) is (-34.1 sec u< u < 264.1 ) seconds.
We are to test the claim that the population mean is greater than 75 sec.
2.
The given confidence interval contains the value of 75 sec, so there is not sufficient evidence to support the claim that the mean is greater than 75 sec.
Read more on confidence interval here:
brainly.com/question/20309162
#SPJ1
Answer:
i am pretty sure it is b
Step-by-step explanation: