Answer:
89
Step-by-step explanation:
<h2>How do you find the range of a data set?</h2><h3><u>first put you're data set from lowest to highest, then subtract your highest by your lowest value</u></h3><h3>Data:</h3>
8, 8, 12, 41, 64, 97
<u>We can clearly see that our highest value is 97, and our lowest in 8.</u>
<h3>97 - 8 = <u>89</u></h3><h2>Hence our range of data is 89.</h2>
Answer: 22.5625
Step-by-step explanation: 4.75^2 squared is the same thing
Answer:
x=6/5 or 1.2
Step-by-step explanation:
I could not tell if your original problem was 2=-5x+8 or if it should have been 2x=-5x+8 so I solved both
2= -5x+8 (next subtract 8 from both sides)
-6= -5x (divide both sides by -5)
-6/-5=x (the negative cancel out next)
6/5 or 1.20 is x
If your original problem should say 2x=-5x+8
2x=-5x+8 (add 5x to both sides next)
7x=8 (divide by 7 on both sides)
x=8/7
Answer:
The answer is attached for better presentation of formulas.
Step-by-step explanation:
The areas under the normal curve between any two ordinates at X=a and X=b equals the probability that the r.v X lies in the interval [a,b]. that is P (a ≤ X ≤ b) = ∫_a^b▒1/(σ√2π) e^((-(x-u))/(2σ^2)) dx
which is represented by the area of the shaded region. (figure1)
But integrals of this type cannot be solved by ordinary means. They are however evaluated by the methods of numerical integration, and numerical approximations for some function have been tabulated for quick reference.
The table of areas under the unit normal curve gives the areas (probabilities) for the standard normal distribution from the mean, z=0 to a specified value of z say z0. Since normal curves are symmetrical therefore P (0 to z) = P (0 to –z). That is why the areas for negative values of z are not tabulated. Hence to use the table of areas for the normal distribution, the values of the r.v X in any problem are changed to the values of the standard normal variable Z and the desired probabilities are obtained from the table.
Thus to find P (a <X<b) we would change X into Z as follows
P (a <X<b) = P ((a-u)/σ ≤ (X-u)/σ ≤ (b-u)/σ )
= P ((a-u)/σ ≤ Z ≤ (b-u)/σ )
Where (a-u)/σ and (b-u)/σ are the z- values of the standard normal variable Z.
In practice, a normal curve sketch for the given problem, showing under the X scale, a scale for the corresponding values of z will help in solving problem. (Figure 2)
Answer:
1.23
2. Question incomplete
3. Needs Question 2 to Answer
Step-by-step explanation:
1. 115÷5=23