What is the equation of the line? Identify the values of A, B, and C in the

Determine the total annual fica tax for an annual salary of $38,480.

egoroff_w [7]

The annual FICA (Federal Insurance Contributions Act) tax is composed of two parts: Social Security and Medicare Taxes. On the employee's side, he/she has to shoulder 6.2% of the annual salary for the Social Security Tax, and 1.45% of the annual salary for the Medicare Tax. So in total, FICA takes 7.65% of your annual salary as tax. To calculate the FICA tax to be paid, this is shown below:

$38,480 x 7.65% = $2943.72

Therefore, for an annual salary of $38,480, $2943.72 goes into FICA tax payment.

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A door delivery florist wishes to estimate the proportion of people in his city that will purchase his flowers. Suppose the true

kvv77 [185]

Answer:

99.74% probability that the sample proportion will be less than 0.1

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .