Answer:
50%
Step-by-step explanation:
68-95-99.7 rule
68% of all values lie within the 1 standard deviation from mean 
95% of all values lie within the 1 standard deviation from mean 
99.7% of all values lie within the 1 standard deviation from mean 
The distribution of the number of daily requests is bell-shaped and has a mean of 55 and a standard deviation of 4.
68% of all values lie within the 1 standard deviation from mean = 
= 
95% of all values lie within the 2 standard deviation from mean = 
= 
99.7% of all values lie within the 3 standard deviation from mean = 
= 
Refer the attached figure
P(43<x<55)=2.5%+13.5%+34%=50%
Hence The approximate percentage of light bulb replacement requests numbering between 43 and 55 is 50%
Step-by-step explanation:
I would go with a or b but one sec let me work it out
Answer:
The correct option is C) The results were not statistically significant but were practically significant.
Step-by-step explanation:
Consider the provided information.
They found that the difference in these shelf lives had a p-value of 0.24. Assume an α of 0.05.
We reject the null hypothesis if p value is less than α.
We are fail to reject null hypothesis if p value is greater or equal to α.
Here p value is greater than 0.05.
So, we do not reject null hypothesis and conclude that result is not statistically significant. But, there is a practical difference between 3 days and 7 days.
Therefore, the correct option is C) The results were not statistically significant but were practically significant.
Answer:
63 cm²
Step-by-step explanation:
the area of ∆YXZ = 7 × 3² = 7×9 = 63 cm²
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