Answer:
Step-by-step explanation:
1.5x^2+6x+.... so that it can be rewritten in the form a(x+b)^2
Let .... be C
1.5x^2+6x+C
=1.5(x^2+4x+(C/1.5))
Compare above with 1.5(x+2)^2 = 1.5(x^2+4x+4)
C/1.5=4
C=6
So the expression should be 1.5x^2+6x+6
It can be rewritten to 1.5(x+2)^2
Using the t-distribution, it is found that the 99% confidence interval for the true difference between the mean distance hit with the new model and the mean distance hit with the older model is (-25.4, 40.4).
We will find the <u>standard deviation for each sample</u>, hence, the t-distribution will be used.
For each sample, the mean, standard deviation and sample sizes are given by:


The standard error for each sample is given by:


The distribution of the differences has <u>mean and standard error</u> given by:


The interval is:

The critical value for a <u>99% two-tailed confidence interval</u>, with 10 + 10 - 2 = <u>18 df</u>, is t = 2.8784.
Then:


The 99% confidence interval for the true difference between the mean distance hit with the new model and the mean distance hit with the older model is (-25.4, 40.4).
To learn more about the use of the t-distribution to build a confidence interval, you can check brainly.com/question/25675821
Answer:20$
Step-by-step explanation:
2x20=40
40+5=45