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timama [110]
2 years ago
7

A pair of pants on sale for 25% off if they're original price was $40 what is the new sale price

Mathematics
1 answer:
notsponge [240]2 years ago
6 0
The answer should be $30 cause it's only 25% off
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Points A, B, C, and D lie on the circle. Solve for the value of X
aliina [53]

Answer:

A 65

Step-by-step explanation:

The angle formed by two chords  is 1/2 the sum of intercepted arcs

x = 1/2 (51+79)

x = 1/2(130)

x =65

4 0
3 years ago
Read 2 more answers
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

s_1 = sample standard deviation for mail-order purchases = $17.25

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.

          = (74.50-84.40) \pm (2.807  \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})

          = [$-31.82 , $12.02]

Hence, 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].

5 0
3 years ago
Oof I am really confused on 13. Can someone help me with it? I’ll really appreciate it. Thanks!
Nataly [62]

Hi! I know for sure that a and d are equivalent to the expression which is not what you are looking for. This is because 11 times 8 gives you the 88 on a and if you multiply the 5 also by 8 you get 40 so a is not the answer. The reason I multiply both by the same number and got the equivalency is because, "What you do to one side, you do to the other." Same thing with d. My answer would be B because you can't change the variable to a different number. You can't just give the five the y and make it mean the same. The answer is B.

3 0
2 years ago
Help please i want help please
Rasek [7]
Don't listen but i'm gonna go with the third one
3 0
3 years ago
56 with 75% increase
valina [46]

Answer:133

Step-by-step explanation:

7 0
2 years ago
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