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Brilliant_brown [7]
3 years ago
9

Mr. Sharon can paper a room in 3hrs. His assistant needs 5hrs. If Mr. Sharon works 1hr and then his assistant completes the job

alone, how long will it take the assistant to finish the job?
Mathematics
1 answer:
stealth61 [152]3 years ago
6 0

4 hours? Sorry if it's wrong

hope I helped!!!

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500 grams of tomatoes cost £1.76 how much will 1.75kg cost??
ValentinkaMS [17]
500 grams equals 0.50 kg so multiply 1.76 3 times to get 5.28, then divide 1.76 by 2 to get .88 add the two together to get 6.16
3 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
Its #4 help please <br> question 4
sasho [114]

Answer:

I think s(2s) = 205

Step-by-step explanation:

4 0
3 years ago
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Which of the following is the missing term of the geometric sequence in which a1= 2 and a3 = 200?
stellarik [79]
A2 = 20.
The formula is a(n+1) = 10a(n)
5 0
3 years ago
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Active peer pressure involves all of the following EXCEPT: A. bribery B. teasing C. put-downs D. famous people
lana [24]

Answer:

d

Step-by-step explanation:

it is d because famous people might do it but it is the lest likey

7 0
3 years ago
Read 2 more answers
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