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Natali [406]
3 years ago
8

15 pounds of beans are distributed equally into 10 bags to give out at the food bank. How many pounds of beans are in each bag?

Enter you answer in simplist form.
Mathematics
1 answer:
Tems11 [23]3 years ago
8 0

Answer:

1.5, or 1 1/2

Step-by-step explanation:

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Find the range. 4.7 6.3 5.4 3.2 4.9 3.1 –3.1 9.5 –9.5
WINSTONCH [101]

Answer:

The range is 19

Step-by-step explanation:

First, you have to order the numbers from least to greatest:

-9.5, -3.1, 3.1, 3.2, 4.7, 4.9, 5.4, 6.3, 9.5

Then, you have to find the difference between the largest number and and smallest number. You do this by subtracting them:

9.5 - (-9.5)= 19

So, the range is 19

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
Kyra see a pair of boots that costs $60.00.They are on salenfor 25% off.She also has a coupon for an additional 10% off.How much
nikitadnepr [17]
It's 40.5
60 \times .25 \\ 45 \times .10
4 0
4 years ago
Nick walked 1 mile in nine minutes there are 5280 feet in a mile about how many feet did he walk a minute
Elodia [21]
Hello!

All you have to do it divide 5280 ft by 9 min  to find ft/min:

5280 / 9 = 506.67 ft/min

I hope this helps :))
3 0
3 years ago
Read 2 more answers
Find the solution set of the system of equations below y=1/2x-1 and x+y=4
DerKrebs [107]
I hope this helps you

4 0
3 years ago
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