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Brilliant_brown [7]
3 years ago
15

⚠️ILL GIVE BRANLIEST!

Mathematics
1 answer:
SSSSS [86.1K]3 years ago
3 0

Answer:

c

Step-by-step explanation:

You might be interested in
Can you make 24 from the numbers 3, 6, 8 and 2, using all the numbers? You can only use the operations: + - x / ( )
Andre45 [30]

Answer:

you can do 3 x 6 + 8 - 2

Step-by-step explanation:

3 × 6= 18

+8= 26

-2= 24

hope this Helps :)

4 0
3 years ago
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
Scientists are creating a material that may replace damaged cartilage in human joints. This hydrogel can stretch to 21 times its
Oduvanchick [21]

Answer:

c

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Sylvia took three times in a video game she scored -120 points during her first turn 320 points in her second time and -80 point
sp2606 [1]
Her average score for the 3 turns is 40 because you have to add all of the scores so -80 and -120 is -200 so add that to 320 and you get 120 divide that by 3 because she took 3 turns and you get 40
3 0
3 years ago
2605 rounded to the nearest hundred
Fittoniya [83]
It would round to 2600
8 0
3 years ago
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