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3 years ago

Answer the following equation: 1/2(x + 20) = 17/4x + 4 (6 - x)

1 answer:

8 0

So use distributive property
which is
a(b+c)=ab+ac so
1/2(x+20)=1/2x+10

and the other
4(6-x)=24-4x

so now we have
1/2x+10=17/4x+24-4x
convert -4x to -16/4x and add to the 17/4x
1/2x+10=1/4x+24
subtract 24 from boths ides
1/2x-14=1/4x
multily both sides by 4
2x-56=x
subtract x from both sides
x-56=0
add 56 to both sides
x=56

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A physical education department wanted a single test of upper body strength that was easy to administer. Dips on the parallel ba

Gennadij [26K]

Answer:

24.23 ; 13.47 ; 0.782 ; p value < 0.00001 ;7.45

Step-by-step explanation:

Given that :

ΣX = 3,416, ΣY = 1,899, SDX = 8.84, SDY = 4.70, ΣZXZY = 109.416

Sample size, N = 141

Mean of X:

£X / N = 3416 / 141 = 24.23

Mean of Y:

£Y / N = 1899 / 141 = 13.47

B.) Correlation between the 2 variables :

£ZXZY / N - 1

= 109.416 / 140

= 0.782

C.)

Level of significance and p value :

T = [Rsqrt(n-2)] ÷ sqrt(1 - R²)

T = [0.782 * sqrt(139)] ÷ sqrt(1-0.782^2)

T = 14.79

P value from Tscore calculator

Df = 141 - 2 = 139 ; α = 0.05

Pvalue = 0.00001

Pvalue < α ; Hence, a significant positive relationship exists.

To obtain regression model :

Slope = £ZXZY / SDX²

m = 109.416 / 8.84^2 = 1. 4

yintercept = meanY - slope*x

yintercept = 13.47 - 1.4*24.3

yintercept = - 20.55

Equation in slope intercept form:

y = mx + c

y = 1.4x - 20.55

For x = 20

y = 1.4(20) - 20.55

y = 7.45

5 0

3 years ago

Rick knows that 1 cup of glue weighs 1/18 pound. he has 2/3 pound of glue. how many cups of glue does he have

Mumz [18]

(2/3) / (1/18) =
2/3 * 18/1 =
36/3 =
12 cups <===

4 0

3 years ago

Read 2 more answers

Find the magnitude of the vector with (-7, -5) and write the vector as the sum of unit vectors.

Lesechka [4]

The magnitude of vector (-7,-5) =
= $\sqrt{ (-7)^{2} + (-5)^{2} } = \sqrt{74} = 8.6$

vector as the sum of unit vectors = -7 i - 5 j

4 0

3 years ago

Read 2 more answers

Find the equation of a line that is perpendicular to the line y=(1/3)x+4 and contains the point (-9,0)?

torisob [31]

Answer:

y = -3x - 27

Step-by-step explanation:

0 = -3[-9] + b

-27 = b

y = -3x - 27

* Perpendicular Lines have OPPOSITE MULTIPLICATIVE INVERSE RATE OF CHANGES [SLOPES]:

⅓ → -3

I am joyous to assist you anytime.

7 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