Find common ratio in the geometric sequence 1,5,25,125
1 answer:
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Penjelasan langkah demi langkah:
1)
2) √32 +3√18-2√50
= √16*2 +3√9*2-2√25*2
= 4√2 + 3(3√2)-2(5√2)
= 4√2 + 9√2-10√2
= 13√2-10√2
= 3√2
3) 1000 ⅔×64⅙
4) 3/4+√2
5) 2√3×√18
= 2√3×√9*2
= 2√3×3√2
= (2*3)(√3*√2)
= 6√6
6) 12/3+√3
= 4+√3
7) √1000—2√40
= 10 -2 (√4*10)
= 10-2(2√10)
= 10 - 4√10
8) 2- ¹+3-¹
9)
Jika pernyataannya opsional, penyebutnya adalah 1
10) 2√3×√18
= 2√3×√9*2
= 2√3×3√2
= (2*3)(√3*√2)
= 6√6
Answer:
88.51 is the minimum score needed to receive a grade of A.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 73
Standard Deviation, σ = 11
We are given that the distribution of exam grades is a bell shaped distribution that is a normal distribution.
Formula:
We have to find the value of x such that the probability is 0.0793.
Calculation the value from standard normal z table, we have,
Hence, 88.51 is the minimum score needed to receive a grade of A.