Can some one plz help me?! The LATERAL Surface Area for the shape below is:
1 answer:
Answer:
430 cm2
Step-by-step explanation:
The Lateral Surface Area of it is equal to:
Area of ABCD + Area of A'B'C'D' + Area of ABB'A' + Area of DCC'D' + Area of ADD'A' + Area of BCC'B'
As ABCDA'B'C'D' is a rectangle prism, so that:
+) Area of ABCD = Area of A'B'C'D'
+) Area of ABB'A' = Area of DCC'D'
+) Area of ADD'A' = Area of BCC'B'
=> Total lateral surface area = 2. Area ABCD + 2. Area ABB'A + 2. Area ADD'A
As ABCDA'B'C'D' is a rectangle prism, so that we have: ABCD, ABB'A', ADD'A' are rectangles.
The formula to calculate the area of rectangle is: Area = Length x Width
We have:
+) Area ABCD = AB x BC =11 x 10 = 110 cm2
+) Area ABB'A = AB x AA' = 11 x 5 = 55 cm2
+) Area ADD'A = AA' x A'D' = 5 x 10 = 50cm2
Total lateral surface area = 2. Area ABCD + 2. Area ABB'A + 2. Area ADD'A
= 2 x 110 + 2 x 55 + 2 x 50 = 430 cm2
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Answer:
The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.
Step-by-step explanation:
We are in posession of the sample's standard deviation, so we use the student's t-distribution to find the confidence interval.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 16 - 1 = 15
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 35 degrees of freedom(y-axis) and a confidence level of ). So we have T = 2.9467
The margin of error is:
M = T*s = 2.9467*0.058 = 0.171
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 1.053 - 0.171 = 0.882kg
The upper end of the interval is the sample mean added to M. So it is 1.053 + 0.171 = 1.224 kg.
The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.