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yarga [219]
2 years ago
14

As a children's librarian, Jodi Hassam earns $15.35 an hour

Mathematics
2 answers:
grandymaker [24]2 years ago
5 0
<h2>Answer:</h2>

Her total pay for the week is:

                      $ 982.40

<h2>Step-by-step explanation:</h2>

It is given that:

Jodi Hassam earns $15.35 an hour  plus time and a half for weekend work.

Last week she worked  her regular 40 hours plus 16 hours on the weekend.

This means that her total pay is calculated as follows:

=40\times 15.35+16\times (1.5\times 15.35)\\\\=982.40

( since the first term represents the pay for the regular hours and the second term represent the pay for the weekend )

Hence, her total pay for the week is: $ 982.40

mihalych1998 [28]2 years ago
3 0

Answer:

Step-by-step explanation:40X15.35+(16(15.35X1.5)= 982.4

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J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.
melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

                      P.Q.  =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }  ~ t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean for mail-order sales = $74.50

\bar X_2 = sample mean for internet sales = $84.40

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s_2 = sample standard deviation for internet purchases = $21.25

n_1 = sample of sales receipts for mail-order purchases = 16

n_2 = sample of sales receipts for internet purchases = 9

Also,  s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }  =  \sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} } = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by (\mu_1-\mu_2).

Now, 99% confidence interval for (\mu_1-\mu_2) is given by;

             = (\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_)  \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

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