Answer:
Interval [16.34 , 21.43]
Step-by-step explanation:
First step. <u>Calculate the mean</u>

Second step. <u>Calculate the standard deviation</u>



As the number of data is less than 30, we must use the t-table to find the interval of confidence.
We have 6 observations, our level of confidence DF is then 6-1=5 and we want our area A to be 80% (0.08).
We must then choose t = 1.476 (see attachment)
Now, we use the formula that gives us the end points of the required interval

where n is the number of observations.
The extremes of the interval are then, rounded to the nearest hundreth, 16.34 and 21.43
Please: Use "^" to denote exponentiation: <span>2x^2 + 8x - 12 = 0
Reduce this by div. every term by 2: </span><span>x^2 + 4x - 6 = 0
Here a=1, b=4 and c = -6. Square half of b, obtaining (4/2)^2 = 4, and add, and then subtract, this 4 to x^2 + 4x - 6:
</span> x^2 + 4x +4 - 4 - 6 = 0. Rewrite the square as (x+2)^2, obtaining new equation
(x+2)^2 = 10. Take the sqrt of both sides: x+2 = plus or minus sqrt(10).
Finally, solve for x: x = -2 plus or minus sqrt(10).
Answer:
203
Step-by-step explanation:
17*7=119
12*7=84
84+119=203
3a + a - 15 = 225 . your substituting b(a-15) into the first equation. <span />