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irina [24]
3 years ago
7

Three robots ran the same distance. The first robot ran 150 meters per minute.

Mathematics
2 answers:
Molodets [167]3 years ago
7 0
Boosting this so more people can see!!
marissa [1.9K]3 years ago
3 0

Answer:

900 meters.

Step-by-step explanation:

Trust me. Russian School Of Math answer :)

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HELP ME PLS W 23 AND 24
n200080 [17]

Answer:

23 is 1/25

24 is -3

Step-by-step explanation:

Please give that brainliest giiirllllll, if you are one thanks tho!!!!!!

add me on snap haha badluckbandit69

8 0
3 years ago
Including 6% sales tax, an inn charges $135.68 per night. Find the inns nightly cost
Andru [333]
135.68(1+0.06)^1= 144
4 0
3 years ago
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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
Determine if the following sequence is an Arithmetic Sequence. If it is, what would be the 31st term? NOTE: an=a1+(n-1)d
Ivan
<h3>Key points :-</h3>

᪥ The formula to find the 31st term is : \sf \: a_{n} =  a_{1} + (n - 1) \times d

᪥ In the formula, \sf \: a_{1} represents the first term of the sequence.

᪥ \sf\:n is the number of terms, In our case n is 31.

᪥ \sf\:d is the common difference between the terms, In our case d is 4.

\green{ \rule{300pt}{3pt}}

<em>Detailed Solution is attached</em>᭄

3 0
2 years ago
Read 2 more answers
a school district requires all graduating seniors to take a mathematics test. This year, the rest scores were approximately norm
Klio2033 [76]
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3 0
3 years ago
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