I believe the answer is D. A credit report covers info on loan paying history. I was stuck between b and d but then realized it said the company gave a loan while your own credit history covers what you have paid or loaned. Hope this helps, have a nice day.
Multiply both sides by -4/7
Hi there,
5x?=110 ?=22
how? 110/5=22
now 2x?=110
110/2=55 ?=55'
so we know now that people bought 22 packages with card and 55 w/th out cards!
so no we need to divide 695/both
695/5=139 packages with cards
695/2=347 packages without cards
SO
139 packages with card
347 packages without card
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:
2
Step-by-step explanation:
2/2+6=7