4x + 11y = 7 4x + 3y = -1

In the service department of the Glenn-Mark Auto Agency, mechanics requiring parts for auto repair or service present their requ

Anton [14]

Answer:

the average time spend is 0.111

Step-by-step explanation:

The computation of the average time spend is as follows;

Arrival rate is

= 210 ÷ 10

= 21 per hour

Now the service rate is

= 60 ÷ service time

= 60 ÷ 2 minutes

= 30 per minute

And, finally the average time spend is

= 1 ÷ (service rate - arrival rate)

= 1 ÷ (30 - 21)

= 0.111

Hence, the average time spend is 0.111

6 0

3 years ago

Find the value of x, given cos 0 = 0.4 PLEASE HELP!!!!!! And explain if possible

elena-14-01-66 [18.8K]

Answer:

$x=30$

Step-by-step explanation:

For each right-angled triangle:

$\cos\theta=\frac{cathetus \ adjacent \ \theta}{hypotenuse}$

Because $\cos \theta=0.4, \ \ cathetus\ adjacent \ \theta \ = \ 12, \ \ \ hypotenuse=x$

we have an equation

$0.4=\frac{12}x$

multiplying both sides by x, we get $0.4x=12$

dividing both sides by 0.4 and $x=\frac{12}{0.4}$

using calculator, we divide 12 by 0.4

and finally x=30

8 0

3 years ago

The College Board SAT college entrance exam consists of three parts: math, writing and critical reading (The World Almanac 2012)

Wittaler [7]

Answer:

Yes, there is a difference between the population mean for the math scores and the population mean for the writing scores.

Test Statistics = $\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$ .

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, $\mu_A$ = Population mean for the math scores

$\mu_B$ = Population mean for the writing scores

Let $\mu_D$ = Difference between the population mean for the math scores and the population mean for the writing scores.

Null Hypothesis, $H_0$ : $\mu_A = \mu_B$ or $\mu_D$ = 0

Alternate Hypothesis, $H_1$ : $\mu_A \neq \mu_B$ or $\mu_D \neq$ 0

Hence, Test Statistics used here will be;

$\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$ where, Dbar = Bbar - Abar

$s_D$ = $\sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}$

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 = $\frac{\sum B_i}{n}$ = $\frac{474+380+463+612+420+526+430+459+615+541+335+613}{12}$ = 489

Abar = $\frac{\sum A_i}{n}$ = $\frac{540+432+528+574+448+502+480+499+610+572+390+593}{12}$ = 514

∑ $D_i^{2}$ = 22600 and $s_D$ = $\sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}$ = $\sqrt{\frac{22600 - 12*(-25)^{2} }{12-1} }$ = 37.05

So, Test statistics = $\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$

= $\frac{-25 - 0}{\frac{37.05}{\sqrt{12} } }$ follows $t_1_1$ = -2.34

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 .

Therefore, we conclude that there is a difference between the population mean for the math scores and the population mean for the writing scores.

8 0

3 years ago

0.134 bar in the form of p/q

kenny6666 [7]

Answer:

67/500

Step-by-step explanation:

we have

The decimal number 0.134

Convert to a fraction number

Multiply by (1,000/1,000) the number

0.134*(1,000/1,000)=134/1,000

Simplify

Divide by 2 both numerator and denominator

67/500

6 0

3 years ago

What is the next number in the pattern below?

ivann1987 [24]

The answer is B.33, this is because the numbers are increasing by 2's. (+2,+4+,+6, ect.)

7 0

3 years ago

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