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den301095 [7]
3 years ago
11

Consider cases where the number of withholding allowances is less than 6. Which will cause the amount of federal income tax with

held to increase?
a.

Gross pay stays the same and withholding allowances increase.


b.

Gross pay increases by more than $20 and withholding allowances stay the same.


c.

Gross pay increases by more than $20 and withholding allowances increase.


d.

Gross pay decreases by more than $20 and withholding allowances stay the same.
Mathematics
2 answers:
KIM [24]3 years ago
8 0

Consider cases where the number of withholding allowances is less than 6.

The amount of federal income tax withheld will increase if the

1. Gross pay increases and withholding allowances remains the same or

2. Gross pay remains the same and withholding allowances decreases

Option B is correct. The amount of federal income tax withheld will increase when Gross pay increases by more than $20 and withholding allowances stay the same

Hope this helps..!!

Thank you :)

Alina [70]3 years ago
8 0

Gross pay increases by more than $20 and withholding allowances stay the same. hope it helps 100% Sure that is correct
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summary(mod)

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lm(formula = y ~ x1 + x2)

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   Min      1Q Median      3Q     Max

-41.730 -12.174   0.791 12.374 40.093

Coefficients:

        Estimate Std. Error t value Pr(>|t|)    

(Intercept) 415.113     82.517   5.031 0.000709 ***  

x1            -6.593      4.859 -1.357 0.207913    

x2            -4.504      1.071 -4.204 0.002292 **  

---  

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1  

Residual standard error: 24.45 on 9 degrees of freedom  

Multiple R-squared: 0.768,     Adjusted R-squared: 0.7164  

F-statistic: 14.9 on 2 and 9 DF, p-value: 0.001395

a).  y=415.113 +(-6.593)x1 +(-4.504)x2

b). s=24.45

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d). F=14.9

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e).  newdata=data.frame(x1=21.3, x2=43)

# confidence interval

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#prediction interval

predict(mod, newdata, interval="predict")

confidence interval

> predict(mod, newdata, interval="confidence",level=.95)

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1 81.03364 43.52379 118.5435

95% CI = (43.52, 118.54)

f).  #prediction interval

> predict(mod, newdata, interval="predict",level=.95)

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1 81.03364 14.19586 147.8714

95% PI=(14.20, 147.87)

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