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Masja [62]
3 years ago
7

Which of the following best describes the gas in a scuba tank? the molecules of gas are moving slowly. the gas is compressed. th

e gas barely fills its container.?
Mathematics
2 answers:
Alekssandra [29.7K]3 years ago
8 0

Answer:

It is compressed

Step-by-step explanation:

I took the test...

artcher [175]3 years ago
5 0
I think the correct answer from the choices listed above would be the third option. The statement that best describes the gas in a scuba tank would be that <span>the gas is compressed. The gas inside is exposed in a very high pressure which compresses it to liquid.</span>
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PLEASE HELP!
bagirrra123 [75]

Answer: c) $30,799.45

Step-by-step explanation:

Assuming there are 52 weeks in a year, the rate and time would be converted to;

Rate = 8%/52 = 8/52%

No. of periods = 9 years * 52 weeks = 468

The amount Mr Mantle would have is;

= 15,000 * ( 1 + 8/52%) ⁴⁶⁸

= $30,799.45276

= $30,799.45

5 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
.......................
Step2247 [10]

Answer:

32 degrees

Step-by-step explanation:

If lines r and t are parallel then it would be 180-148

3 0
3 years ago
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Elenna [48]

Answer:

Step-by-step explanation:

Substituting x and y values into the equations, we can get the equation

y = 7 -3x

slope: -3

8 0
3 years ago
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Vera_Pavlovna [14]

Answer:

- 1

Step-by-step explanation:

Slope of the line

=  \frac{3 - 4}{5 - 4}  \\  \\  =  \frac{ - 1}{1}  \\  \\  =  - 1

7 0
3 years ago
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