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Len [333]
3 years ago
5

8. FREEBIE

Mathematics
2 answers:
Pie3 years ago
6 0
C millimeters hope it helps
swat323 years ago
4 0

Answer:

C. milliliter

Step-by-step explanation:

What is the appropriate metric unit for the capacity of a bottle of ink?

See Answer

milliliters.

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Which statement correctly compares the two functions?
d1i1m1o1n [39]

Answer:

you forgot the picture

Step-by-step explanation:

7 0
3 years ago
Solve: − &lt;− + 7( − 2) <br> (show your work)
Rashid [163]

Answer:

hope it helps.

<h2>stay safe healthy and happy...</h2>

3 0
3 years ago
Harris Teeter is selling gala apples for 0.98 per pound Adam purchase 3.5 pounds of the apples how much did she pay for her purc
Schach [20]
Total purchase = (unit cost)(# of pounds purchased)

Here, that total is ($0.98/lb)(3.5 lb) = $3.43 (answer)
8 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
Find two numbers that up to 158. The greater number is 40 more than the lesser number.
snow_tiger [21]
99 and 59

X=big number
X-40 = small

X+ x-40 =158
2X -40=158
2X =198
X=99. Big number is 99
99-40 =59

99+59=158
3 0
3 years ago
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