We have a sample of 28 data points. The sample mean is 30.0 and the sample standard deviation is 2.40. The confidence level required is 98%. Then, we calculate α by:
The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:
![CI(\mu)=\lbrack x-Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}},x+Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}}\rbrack](https://tex.z-dn.net/?f=CI%28%5Cmu%29%3D%5Clbrack%20x-Z_%7B1-%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ccdot%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%5B%5D%7Bn%7D%7D%2Cx%2BZ_%7B1-%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ccdot%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%5B%5D%7Bn%7D%7D%5Crbrack)
Where n is the sample size, and Z is the z-score for 1 - α/2. Using the known values:
![CI(\mu)=\lbrack30.0-Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}}\rbrack](https://tex.z-dn.net/?f=CI%28%5Cmu%29%3D%5Clbrack30.0-Z_%7B0.99%7D%5Ccdot%5Cfrac%7B2.40%7D%7B%5Csqrt%5B%5D%7B28%7D%7D%2C30.0%2BZ_%7B0.99%7D%5Ccdot%5Cfrac%7B2.40%7D%7B%5Csqrt%5B%5D%7B28%7D%7D%5Crbrack)
Where (from tables):
Finally, the interval at 98% confidence level is:
Answer:
x2 - 10x - 1994
Step-by-step explanation:
Step 1 :
Trying to factor by splitting the middle term
1.1 Factoring x2-10x-1994
The first term is, x2 its coefficient is 1 .
The middle term is, -10x its coefficient is -10 .
The last term, "the constant", is -1994
Step-1 : Multiply the coefficient of the first term by the constant 1 • -1994 = -1994
Step-2 : Find two factors of -1994 whose sum equals the coefficient of the middle term, which is -10 .
-1994 + 1 = -1993
-997 + 2 = -995
-2 + 997 = 995
-1 + 1994 = 1993
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Final result :
x2 - 10x - 1994
Processing ends successfully
plz mark me as brainliest :)
1( n * 2)
2(n+2*x)
i hope this helps the question wasn't very specific
8m,3n,9p would im guessing make the numbers to the rest of the problem
