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Katyanochek1 [597]
3 years ago
6

A classmate says that a rectangular prism that is 6 cm long, 8 cm wide, and 15 cm high is similar to a rectangular prism that is

12 cm long, 14 cm wide, and 21 cm high. Explain your classmate's error.
Mathematics
1 answer:
Elza [17]3 years ago
8 0
I think you could try to multiply all three of the first prism and then multiply all three of the second prism. Then you can compare both of them to each other, and there will be a difference. 
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Alina [70]

Answer: Expression I & Expression II

Step-by-step explanation:

2n+n+3n+5 simplifies to 6n+5

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Part 2: Create a scenario for a geometric sequence. For example, Anthony goes to the gym for ______ minutes on Monday. Every day
Ugo [173]
An example scenario is:

Anthony goes to the gym for <em>20</em> minutes on Monday. Every day he <em>multiplies</em> his gym time by <em>2</em>. 

On the fifth day, he will spend spend 320 minutes in the gym. 

The formula used to determine the 5th term is,

                                       a5 = 20 x 2^(r -1) 

where r is the common ratio equal to 2. 


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3 years ago
Help! I need the answers quick please
Norma-Jean [14]

Answer:

No idea

Step-by-step explanation:

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3 years ago
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Viktor [21]

Answer:

30 ft²

Step-by-step explanation:

Area of trapezium = 1/2 (a + b)h

a = 6 ft , b = 9 ft , h = 4 ft

Area = 1/2 (6 + 9) 4

= 30 ft²

6 0
2 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
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