Answer:
The confidence interval is from 69.82 o 82.18
Step-by-step explanation:
Using this formula X ± Z (s/√n)
Where
X = 76 --------------------------Mean
S = Standard Deviation
If Variance = 144
S = √144
S = 12
n = 25 ----------------------------------Number of observation
Z = 2.576 ------------------------------The chosen Z-value from the confidence table below
Confidence Interval || Z
80%. || 1.282
85% || 1.440
90%. || 1.645
95%. || 1.960
99%. || 2.576
99.5%. || 2.807
99.9%. || 3.291
Substituting these values in the formula
Confidence Interval (CI) = 76 ± 2.576 (12/√25)
CI = 76 ± 2.576(12/5)
CI = 76 ± 2.576(2.4)
CI = 76 ± 6.1824
CI = 76 + 6.1824 ~ 76 - 6.1824
CI = 82.1824 ~ 69.8176
CI = 82.18 ~ 69.82
In other words the confidence interval is from 69.8176 to 82.1824
Answer:
A. 0.15
B. 0.45
C. 85
Step-by-step explanation:
Given that:
Equation to represent the relation between distance, 
in miles and time 
in seconds:
<em>A. </em>
To find: The speed of airplane in miles per second.
Formula for speed is given as:
Here, we will have to find the value of 
.
Dividing the given equation by , we get:
is in miles and is in seconds.
Therefore, the answer is <em>0.15</em>.
<em>B.</em>
Distance traveled in 30 seconds.
Given that: = 30 seconds
To find: 
in miles
<em>C.</em>
The time taken to travel a distance of 12.75 miles.
Given that : 
To find: 
seconds
125,594 miles. Canada coastline (x) - U.S. Coastline (12,383) = 113,211.
x - 12,383 = 113,211
x - 12,383 + 12,383 = 113,211 + 12,383 (by addition property of equality)
x = 113,211 + 12,383
x = 125,594
Basically, you add the two given numbers together.
Answer:
A standard normal distribution refers to a normal distribution with a mean of 0 and a standard deviation of 1. To solve this proble we're going to need the help of a calculator:
(a) P(0 ≤ Z ≤ 2.38) = 0.4913
(b) P(0 ≤ Z ≤ 1) = 0.3413
(c) P(−2.70 ≤ Z ≤ 0) = 0.4965
(d) P(−2.70 ≤ Z ≤ 2.70) = 0.9931
(e) P(Z ≤ 1.62) = 0.9474
(f) P(−1.55 ≤ Z)= 0.9394
(g) P(−1.70 ≤ Z ≤ 2.00) = 0.9327
(h) P(1.62 ≤ Z ≤ 2.50) = 0.0464
(i) P(1.70 ≤ Z) = 0.0445
(j) P(|Z| ≤ 2.50) = 0.9876
All values are verified! ✅
