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Anastaziya [24]
3 years ago
10

The graph shows how the volume of a gas sample changes as the temperature changes and the pressure remains constant. What is the

rate of change of the volume of the gas sample with respect to the temperature?

Mathematics
1 answer:
makkiz [27]3 years ago
5 0

Answer:

Assumming the y intercept is at 0,20, the slope is \frac{1}{3}

Step-by-step explanation: \frac{36-26}{50-20} = \frac{10}{30} = \frac{1}{3}

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Help pls I’m #######} my teacher gave this to me without knowing nun
butalik [34]

Basically, you just move every individual point up, down, left, or right by the amount indicated.

For example, point G is graphed on the point (-3, -1). Moving it right 5 and up 1 will give you G’, which is (2,0)

T is graphed on the point (-1, -1). Moving it right 5 and up 1 will give you T’, which is (4,0)

B is graphed on the point (-3, -5). Moving it right 5 and up 1 will give you B’, which is

(2,4)

7 0
3 years ago
A market research group is interested in comparing the mean weight loss for two different popular diets. The researcher chooses
Degger [83]

Answer:

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Step-by-step explanation:Jawo

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
Net income is equal to the total revenue from selling an item minus the cost of producing the item. Determine the net income if
Travka [436]
C-Cost because it the most reasonableness one
3 0
3 years ago
Roy's average balance checking account pays simple interest of 4.8% annually, and he made $2.25 in interest last month. What was
Klio2033 [76]

The first thing you do is you divide the annual interest into 12 months

4.8 % / 12

Make sure that you convert the interest

So 0.048/12

Then you get the interest last month so that is

2.25/ (0.048/12)

The answer is 562.5

B.$562.5

5 0
3 years ago
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