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Evgesh-ka [11]
2 years ago
14

A rectangle has an area of 270m2.

Mathematics
1 answer:
Papessa [141]2 years ago
8 0
X * 3 = 270
Solving for x, x = 90

So one side is 90m and the other is 3m.
Now find entire perimeter 90+90+3+3=180+6=186m
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The mass of a rock sample is 25 g and
balu736 [363]

Answer:

given us,

mass= 25g

density= 5 g/ cm2

here,

volume = mass/ density

= 25g/ 5g cm2

= 5cm2

Step-by-step explanation:

cm2 is the square

3 0
3 years ago
Read 2 more answers
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
3) There were 35 children and 10 adults at a barbeque.
Mila [183]

Answer:

3. a. 10:35= 2:7

b.35:45=7:9

c.40:50=4:5

7 0
3 years ago
What is the area of this figure?<br><br> Enter your answer in the box.
Hoochie [10]

Answer:

6 i think double check though

Step-by-step explanation:

3 0
3 years ago
The train to Boston traveled at an average speed of 50 miles per hour for 5 hours. The train to Hartford traveled at an average
lawyer [7]
To find how many miles per hour you need to divide 50 by 5 to get the number of miles per hour and do the same thing to the 2nd step and to find the difference you need to subtract the numbers. Hope it helps!!!
5 0
3 years ago
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