This question please.

nevsk [136]

Answer:

Answer of 17 is ㏒( $x^{2}$ +15x), Answer of 33 is x = 8 , Answer of 35 is x = ㏒10/㏒2 , Answer of 37 is x = -㏒12/㏒8 and Answer of 39 is x = 5

Step-by-step explanation:

17. ㏒x + ㏒(x+15)

Using property ㏒a + ㏒b = ㏒a×b

∴ ㏒x + ㏒(x+15)

㏒x×(x+15)

㏒( $x^{2}$ +15x)

The answer is ㏒( $x^{2}$ +15x)

33. 2^(x-5) = 8

2^(x-5) = 2^3

Using property 2^a = 2^b

Then a = b

∴x-5 = 3

x = 8

The answer is x = 8

35. 2^x = 10

Taking log on both sides gives

㏒2^x = ㏒10

x×㏒2 = ㏒10

x = ㏒10/㏒2

The answer is x = ㏒10/㏒2

37. 8^-x = 12

Taking log on both sides gives

㏒8^-x = ㏒12

-x×㏒8 = ㏒12

x = -㏒12/㏒8

The answer is x = -㏒12/㏒8

39. 5(2^3 × x) = 8

5(8×x) = 8

x = 5

The answer is x=5

5 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

The function h defined by h(t)=(49 + 4.9t)(10 - t) models the height, in meters, of an object t seconds after it is dropped from

kipiarov [429]

Answer:

Step-by-step explanation:

Let's FOIL this out and get it into standard quadratic format:

$h(t)=-4.9t^2+490$ . The lack of a linear term in the middle means therewas no upwards velocity, consistent with the object being dropped straight down as opposed to thrown up in the air tand then falling in a parabolic path. The -4.9t² represents the acceleration due to gravity, and the 490 represents the height from which the object was dropped. The constant in a quadratic that is modeling parabolic motion always represents the height from which the object was dropped (or launched). That's how you know.