Ruben sees 14 wheels on a total of 6 bicycles and tricycles. How many bicycles and tricycles are there?
2 answers:
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Using a system of equation, we have that:
a)
- x, which is the cost of on-site service.
- y, which is the cost of at-store service.
- z, which is the cost of by-mail service.
b)
The system is:
c)
The cost of one on-site repair service is of $90.
Item a:
The variables are:
- x, which is the cost of on-site service.
- y, which is the cost of at-store service.
- z, which is the cost of by-mail service.
Item b:
On-site service costs <u>3 times as much as at-store service</u>, thus:
By mail service costs <u>$10 less than at-store</u> service, thus:
Last week, the shop completed <u>15 services on-site, 40 services at-store, and 5 services by mail for total sales of $3100</u>, thus:
The system is:
Item c:
- The cost of one on-site repair service is x.
- First, replacing the first two equations into the third, we find y, and then with it we find x.
Then
The cost of one on-site repair service is of $90.
A similar problem is given at brainly.com/question/24823220
Answer:
Yes, there is a difference between the population mean for the math scores and the population mean for the writing scores.
Test Statistics = follows
.
Step-by-step explanation:
We are provided with the sample data showing the math and writing scores for a sample of twelve students who took the SAT ;
Let A = Math Scores ,B = Writing Scores and D = difference between both
So, = Population mean for the math scores
= Population mean for the writing scores
Let = Difference between the population mean for the math scores and the population mean for the writing scores.
<em> Null Hypothesis, </em><em> : </em>
<em> or </em>
<em> = 0 </em>
<em> Alternate Hypothesis, </em><em> : </em>
<em> or </em>
<em> 0</em>
Hence, Test Statistics used here will be;
follows
where, Dbar = Bbar - Abar
=
n = 12
Student Math scores (A) Writing scores (B) D = B - A
1 540 474 -66
2 432 380 -52
3 528 463 -65
4 574 612 38
5 448 420 -28
6 502 526 24
7 480 430 -50
8 499 459 -40
9 610 615 5
10 572 541 -31
11 390 335 -55
12 593 613 20
Now Dbar = Bbar - Abar = 489 - 514 = -25
Bbar = =
= 489
Abar = =
= 514
∑ = 22600 and
=
=
= 37.05
So, Test statistics = follows
= follows
= -2.34
<em>Now at 5% level of significance our t table is giving critical values of -2.201 and 2.201 for two tail test. Since our test statistics doesn't fall between these two values as it is less than -2.201 so we have sufficient evidence to reject null hypothesis as our test statistics fall in the rejection region .</em>
Therefore, we conclude that there is a difference between the population mean for the math scores and the population mean for the writing scores.