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anastassius [24]
3 years ago
7

I’ll mark brainiest please help

Mathematics
1 answer:
kicyunya [14]3 years ago
4 0

Answer:

idrk

Step-by-step explanation:

You might be interested in
What is the surface area of a sphere with a radius of 11 units
Otrada [13]

Answer: A≈1519.76³ Units

Step-by-step explanation:

A=4πr²

r = 11

11²=121

121 * 3.14 = 379.94

379.94(4) = 1519.76

A≈1519.76³ Units OR

A=4πr2=4·π·112≈1520.53084

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
Mia's soccer coach told her to keep track of how much water she drinks. Yesterday she drank 18 ounces with breakfast, 21 ounces
igor_vitrenko [27]
Yes, 82 ounces is the correct answer.
6 0
3 years ago
True or False?
dsp73

Answer:

true

Step-by-step explanation:

 i go to k12 i did this test

6 0
2 years ago
A new car is purchased for 23700 dollars. The value of the car depreciates at 13.5% per year. What will the value of the car be,
Marianna [84]
The answer is 23,698.65 : )
7 0
3 years ago
Read 2 more answers
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