Answer:
B) 103.25
Step-by-step explanation:
Area of rectangle:
A = bh
A = (16)(4)
A = 64
Area of a semi-circle:
r = d/2 = 5
A = (πr²)/2
A = (3.14 * 5²)/2
A = (3.14 * 25)/2
A = (78.5)/2
A = 39.25
Combined:
64 + 39.25 = 103.25
Answer:
In order to subtract a whole number from a fraction, you first have to turn the hole number into a mixed fraction.
Step-by-step explanation:
Say your fraction was 2/3 and your whole number was 5.
They have to have like denominators
5/1 - 2/3 = 15/3 - 2/3
15/3 - 2/3 = 13/3
13/3 = 4 1/3
If scores on an exam follow an approximately normal distribution with a mean of 76.4 and a standard deviation of 6.1 points, then the minimum score you would need to be in the top 2% is equal to 88.929.
A problem of this type in mathematics can be characterized as a normal distribution problem. We can use the z-score to solve it by using the formula;
Z = x - μ / σ
In this formula the standard score is represented by Z, the observed value is represented by x, the mean is represented by μ, and the standard deviation is represented by σ.
The p-value can be used to determine the z-score with the help of a standard table.
As we have to find the minimum score to be in the top 2%, p-value = 0.02
The z-score that is found to correspond with this p-value of 0.02 in the standard table is 2.054
Therefore,
2.054 = x - 76.4 ÷ 6.1
2.054 × 6.1 = x - 76.4
12.529 = x - 76.4
12.529 + 76.4 = x
x = 88.929
Hence 88.929 is calculated to be the lowest score required to be in the top 2%.
To learn more about normal distribution, click here:
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Answer:
2
Step-by-step explanation:
1+1=2
Answer: A. "Segment AD bisects angle CAB." is the right answer.
Step-by-step explanation:
Given : In ΔABC ,AC≅AB.
⇒∠ACB=∠CBA....(1) (∵ angles opposite to equal sides of a triangle are equal )
Now in ΔACD and ΔABD
AD=AD (common)....(2)
Here we need one more statement to prove the triangles congruent that is only statement (A) fits in it.
If AD bisects ∠CAB then ∠CAD=∠BAD..(3)
Now again Now in ΔACD and ΔABD
∠ACB=∠CBA [from (1)]
AD=AD [common]
∠CAD=∠BAD [from (3)]
So by ASA congruency criteria ΔADC≅ΔABD.
