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Anastasy [175]
2 years ago
12

Commission: 10 percent on first $5,000; 15% over $5,000. Find the total graduate commission on $22,000.

Mathematics
1 answer:
natta225 [31]2 years ago
8 0

Based on the information about the percentage, the commission that will be paid will be $3050.

<h3>How to solve percentage</h3>

From the information given, it was stated that there's a 10 percent on first $5,000 and 15% over $5,000.

The total graduate commission on $22,000 will be:

= (10% × $5000) + (15% × $17000)

= (0.1 × $5000) + (0.15 × $17000)

= $500 + $2550

= $3050

Learn more about percentages on:

brainly.com/question/24304697

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Answer:

3x + 9

General Formulas and Concepts:

<u>Pre-Algebra</u>

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Step-by-step explanation:

<u>Step 1: Define</u>

(5x + 3) - (2x - 6)

<u>Step 2: Simplify</u>

  1. Distribute negative:                    5x + 3 - 2x + 6
  2. Combine like terms (x):               3x + 3 + 6
  3. Combine like terms (Z):              3x + 9
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Andre's band is made up of 6 members. Thete are 3 guitar players, 1 drummer, and 2 singers. What id the ratio od drummers to sin
gogolik [260]

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C. 1/2

Step-by-step explanation:

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In a study conducted by the US department of Health and Human Services, a sample of 546 boys aged 6-11 was weighed, and it was d
saveliy_v [14]

Answer:

a) CI = ( - 0,0087 ; 0,0355)

b) No CI contains 0 meaning that statistically the proportions could be equal

Step-by-step explanation:

Boys Sample

Size sample   n₁  = 546

x₁ = 87

proportion   p₁  = 87/546     p₁ = 0,159       p₁  =  15,9 %

Girls sample

Size sample   n₂  = 508

x₂ = 74

proportion  p₂  =  74 / 508     p₂ = 0,1456   p₂  = 14,56 %

Hypothesis test:

Null hypothesis                    H₀              p₁  =  p₂

Alternative Hypothesis       Hₐ              p₁  ≠  p₂

Confidence Interval  CI = 90 %   significance level  α = 10 %

α = 0,1    and as the alñternative hypothesis indicates is a two-tail test

then α/2 = 0,05

z(score)  for 0,05 is from z table   z(c) = 1,64

Confidence Interval 90 % is:

CI = [( p₁ - p₂) ± z(c) *  √ p*q * ( 1/n₁  +  1 / n₂ )

p = (x₁ + x₂ ) / n₁ + n₂

p = ( 87 + 74 ) / 546 + 508

p = 0,1527      then   q = 1 - 0,1527    q = 0,8473

CI = 0,0134 ± √ 0,1527*0,8473 ( 1/546   + 1 / 508 )

CI = 0,0134 ± 0,0221

CI = ( - 0,0087 ; 0,0355)

CI  contains  0 meaning that difference between the groups could be 0

Therefore we can conclude that  proportion on both groups are different. We can´t reject H₀

We need to calculate z(s)

z(s) = [ p₁ - p₂ ] / √ p*q * ( 1/n₁  +  1 / n₂ )

p = (x₁ + x₂ ) / n₁ + n₂

p = ( 87 + 74 ) / 546 + 508

p = 0,1527      then   q = 1 - 0,1527    q = 0,8473

z(s) = [ 0,159 - 0,1456] / √ 0,1527*0,8473 ( 1/546   + 1 / 508 )

z(s) = 0,0134/ √0,1294 ( 0,0018 + 0,00196 )

z(s) = 0,0134/ √0,1294 ( 0,0038)

z(s) = 0,0134/ 0,0221

z(s) = 0,61

p-value for  z(s) = 0,61      p-value = 0,7291

p-value > 0,05

As p-value is bigger than 0,05 we have to accept H₀. We don´t have enough evidence to claim any difference between the two groups

6 0
3 years ago
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