Answer:
Step-by-step explanation:
Given that :
μ1 = 39420 ; σ1 = 1659 ; s1 = 12
μ2 = 30215 ; σ2 = 4116 ; s2 = 26
df1 = 12 - 1 = 11
df2 = 26 - 1 = 25
98 % confidence interval of the difference in sample means :
Pooled Variance :
S²p = ((df1 * s1²) + (df2 * s2²)) / (df1 + df2)
((11×1659^2)+(25×4116^2))÷(11+25) = 12605874.75
Standard Error :
√(S²p/n1) + (S²p/n2)
√(12605874.75/12) + (12605874.75/26)
= 1239.0847
The answer to this question would be: $1888.33/semi-month
To answer this question you need to convert the annual into semi-monthly. Annual mean every 1 year or every 12 months. Semi-monthly mean two times a month. Then the equation would be: $45,320 / year x (1 year/12month) x (1 month/2 semi-month) = $1888.33/semi-month
His usual straight-time pay is $603.327375 per week
<h3>How to determine his usual straight-time pay?</h3>
The given parameters are
Rate = $16.655 per hour
Duration = 36.225 hours per week
His usual straight-time pay is calculated as
Usual pay = Rate * Duration
So, we have
Usual pay = $16.655 per hour * 36.225 hours per week
Evaluate the product
Usual pay = $603.327375 per week
Hence, his usual straight-time pay is $603.327375 per week
Read more about linear rates at
brainly.com/question/7040405
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Hello,
x=R*cos(t)
y=R*sin(t)
y=2x+1
==> Rsin(t)=2Rcos(t)+1
==>R(sin(t)-2cos(t))=1
==>R=1/(sin(t)-2cos(t))
