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Nes M.
Enrichment Teacher, Substitute Teacher, Tutor for 6 years
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Pre-Algebra
TutorMe
Question:

Solve for : 4x =12

Nes M.
Answer:

4(x) = 12 4(3)=12 12=12

Statistics
TutorMe
Question:

What two sample means enclose the central [(1-α*100]% of the distribution of means for conscientiousness (µ = 3.45 and σ = .72) when sample size is n = 36 and α = .05?

Nes M.
Answer:

Step 1 find Z on the Z table zα/2 = 1.96. Step 2 Find Standard Deviation σm = 𝜎/𝑛 = .72/Squared root of36 = .12 Step 3 Find μm ± zα/2 (σm) = 3.45 ± 1.96(.12) = 3.45 ± .24 = [3.21, 3.69]. The interval [3.21, 3.69] encloses the central 95% of a distribution of means with μm = 3.45 and σm = .12

Algebra
TutorMe
Question:

Write 2 log3 x + log3 5 as a single logarithmic expression.

Nes M.
Answer:

log3 x2 + log3 5 = log3(5x2)

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