I believe it would be 6(10^4)
Answer:
Step-by-step explanation:
Given that
Group Group One Group Two
Mean 26.00 23.00
SD 2.00 4.00
SEM 0.40 1.21
N 25 11
where group I represents female servers and group II male servers.
We have to calculate confidence interval for 90% for difference in means
The mean of Group One minus Group Two equals 3.00
df = 34
standard error of difference = 0.993
t critical = 2.034 for 90% df 34
Hence confid. interval at 90%
=Mean diff ±2.034 * std error of diff
= (0.98, 5.02)
Answer:
There is a scale factor of 1
Step-by-step explanation:
the ratio 1:1 means that there is no difference between the two triangles, and that they are congruent.
Complete question :
Standardized tests: In a particular year, the mean score on the ACT test was 19.3 and the standard deviation was 5.3. The mean score on the SAT mathematics test was 532 and the standard deviation was 128. The distributions of both scores were approximately bell-shaped. Round the answers to at least two decimal places. Part: 0/4 Part 1 of 4 (a) Find the z-score for an ACT score of 26. The Z-score for an ACT score of 26 is
Answer:
1.26
Step-by-step explanation:
Given that:
For ACT:
Mean score, m = 19.3
Standard deviation, s = 5.3
Zscore for ACT score of 26;
Using the Zscore formula :
(x - mean) / standard deviation
x = 26
Zscore :
(26 - 19.3) / 5.3
= 6.7 / 5.3
= 1.2641509
= 1.26
