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1 month ago

Complete the squares so that the differences shown on the outside are correct

Mathematics

2 answers:

zzz [12.3K]1 month ago

4 0

8 5 and 18 8

7 2 9 2

The differences calculated are 8 - 5 = 3, 8 - 7 = 1, 5 - 2 = 3, and for the other set, 18 - 8 = 10, 18 - 9 = 9, 9 - 2 = 7, 8 - 2 = 6

AnnZ [12.3K]1 month ago

3 0

Response:

8-1=7

8-3=5

7-6=1

10-9=1

Detailed explanation:

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How much of a 90% alloy must be combined with 70% gold alloy in order to get 60 ounces of an 85% gold alloy?

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24 days ago

Read 2 more answers

At a recent county fair, you observed that at one stand people's weight was forecasted, and were surprised by the accuracy (with

Leona [12618]

Answer:

a) Slope: $\hat \beta_1 =\frac{7625.9}{1248.9}=6.106$

Intercept: $\hat \beta_o = 157.955 -6.106 (69.686)=-267.548$

b) $r=\frac{7625.9}{\sqrt{[1248.9][94228.8]}}=0.657$

Additionally, the coefficient of determination is $r^2 = 0.657^2 =0.432$

Step-by-step explanation:

Definitions and data provided

The correlation coefficient is a measure that quantifies the strength of the relationship between two variable movements, denoted as r and ranging between -1 and 1.

The sum of squares refers to the total of the squared deviations, where deviation is defined as the difference between each value and the grand mean.

When performing multiple regression, the aim is to analyze the relationship between multiple independent variables and one dependent variable.

n=110, $\sum x_i y_i = \sum (X-\bar X)(Y-\bar Y) =7625.9,\sum x^2_i=\sum (x-\bar x)^2 =1248.9, sum y^2_i=\sum(y-\bar y)^2 =94228.8$

$\sum Y_i =17375, \sum X_i = 7665.5$

Part a

The slope can be calculated using this formula:

$\hat \beta_1 =\frac{\sum (x-\bar x) (y-\bar y)}{\sum (x-\bar x )^2}$

Following the substitutions, we have:

$\hat \beta_1 =\frac{7625.9}{1248.9}=6.106$

The intercept can be determined with this formula:

$\hat \beta_o = \bar y -\hat \beta_1 \bar x$

Average values for x and y can be calculated this way:

$\bar X=7665.5/110 =69.686, \bar y= 17375/110=157.955$

Replacing yields:

$\hat \beta_o = 157.955 -6.106 (69.686)=-267.548$

Part b

The correlation coefficient can be calculated using the following formula:

$r=\frac{\sum (x-\bar x)(y-\bar y) }{\sqrt{[\sum (x-\bar x)^2][\sum(y-\bar y)^2]}}$

In our situation:

n=110, $\sum x_i y_i = \sum (X-\bar X)(Y-\bar Y) =7625.9,\sum x^2_i=\sum (x-\bar x)^2 =1248.9, sum y^2_i=\sum(y-\bar y)^2 =94228.8$

We can compute the correlation coefficient by substituting values:

$r=\frac{7625.9}{\sqrt{[1248.9][94228.8]}}=0.657$

The coefficient of determination is $r^2 = 0.657^2 =0.432$

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1 month ago

What is the midpoint of a line segment with the endpoints (-6, -3) and (9, -7)?

PIT_PIT [12445]

The midpoint of the line segment with endpoints (-6, -3) and (9, -7) is (1.5,-5).

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1 month ago

What is the logarithm of the equilibrium constant, log K, at 25°C of the voltaic cell constructed from the following two half-re

lawyer [12517]

Answer:

4.0921 reflects the logarithm of the equilibrium constant.

Step-by-step explanation:

$Fe^{2+} (aq) +2e^{-}\rightarrow Fe(s)$ ; E° = - 0.41 V

$Ag^+(aq) + e^-\rightarrow Ag(s)$ ; E° = 0.80 V

Iron has a negative reduction potential, indicating its tendency to lose electrons and undergo oxidation, and thus it will be at the anode.

$E^{o}_{cell}$ =Reduction potential of cathode - Reduction potential of anode

$E^{o}_{cell}=E^{o}_c-E^{o}_a$

$=0.80 V-(-0.41 V)=1.21 V$

$Fe^{2+} (aq) + 2e^{-}\rightarrow Fe(s)$ ; E° = - 0.41 V

$2Ag^+(aq) + 2e^-\rightarrow 2Ag(s)$ ; E° = 0.80 V

Net reaction: $Fe(s)+2Ag^{+}\rightarrow Fe^{2+}+2Ag(s)$

n = 2

To determine the equilibrium constant, we utilize the correlation with Gibbs free energy, as follows:

$\Delta G^o=-nfE^o_{cell}$

and,

$\Delta G^o=-RT\ln K_{eq}$

Aligning these two equations yields:

$nfE^o_{cell}=RT\ln K_{eq}$

where,

n = electrons transferred = 2

F = Faraday's constant = 96500 C

$E^o_{cell}$ = standard electrode potential of the cell = 1.21 V

R = gas constant = 8.314 J/K.mol

T = reaction temperature = $25^oC=[273+25]=298K$

Substituting values into the equation, we arrive at:

$2\times 96500\times 1.21 V=8.314\times 298\times \ln K_{eq}$

$\ln K_{eq}=9.3478$

$\log K_{eq}=\frac{9.3478}{2.303}=4.0921$

4.0921 represents the logarithm of the equilibrium constant.

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