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valentinak56 [21]
3 years ago
12

2. Casey says, "I pick a number.

Mathematics
1 answer:
victus00 [196]3 years ago
4 0

Step-by-step explanation:

7,14,21,28,35,42,49,56,63,70

18,36,54,72,90,108,126,144,162,180

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O7v2 hope that helps ya! Mark as brainlest

6 0
2 years ago
Solve the following equation: |7 – (24 ÷ | 3-6 |)|<br>(A)15(B)-12(C)1(D)-1
Zepler [3.9K]
PEDMAS

The rule is Paranthesis First

|7-(24÷|3-6|)|

|7-(24÷|-3|)|

Now we have to get the -3 outside the 2 Lines but the lines are saying that if the number inside is negative when you get it outside it has to be positive.

|7-(24÷3)|-

|7-(8)|

|7-8|

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3 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
How can a histogram help you represent a large data set
Inessa05 [86]

Answer:  a histogram provides a visual representation of data distribution. Histograms can display a large amount of data and the frequency.  We can use it to get the frequency of values in a dataset.

6 0
3 years ago
Read 2 more answers
11/5,0.05,5,0.5,1/5 from least to greatest
anygoal [31]

0.05,1/5,0.5,11/5,5                                                                      

3 0
3 years ago
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