18.6 rounded to nearest tenth is 19.0 or just 19
Answer:
{58.02007 , 61.97993]
Step-by-step explanation:
Data are given in the question
Sample of cars = n = 121
Average speed = sample mean = 60
Standard deviation = sd = 11
And we assume
95% confidence t-score = 1.97993
Therefore
Confidence interval is
![= [60 - \frac{1.97993 \times 11}{\sqrt{121} }] , [60 + \frac{1.97993 \times 11}{\sqrt{121} }]](https://tex.z-dn.net/?f=%3D%20%5B60%20-%20%20%5Cfrac%7B1.97993%20%5Ctimes%2011%7D%7B%5Csqrt%7B121%7D%20%7D%5D%20%2C%20%20%5B60%20%2B%20%20%5Cfrac%7B1.97993%20%5Ctimes%2011%7D%7B%5Csqrt%7B121%7D%20%7D%5D)
= {58.02007 , 61.97993]
Basically we applied the above formula to determine the confidence interval
The equation would be y=4/3x+3.
The number that comes before x is the slope, and the number that is after the + is where it crosses the y axis.
Hope this helped! (Can I maybe have brainliest?)
12.5 is already a decimal...
If you mean 12/5, then it is 2 and 2/5, which is 2.4.
If you mean 12 x 5, then it's 60.
If you mean fractions equivalent to 12.5 then it's 25/2.
If you mean decimals then 12.50, or 12.500.
Hope this helps