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3 years ago

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Mathematics

1 answer:

Sunny_sXe [5.5K]3 years ago

6 0

The solutions of the equations are -1.3 and -7.7

Step-by-step explanation:

The quadratic formula of solving the quadratic equation

ax² + bx + c = 0, where a, b and c are constant is:

$x=\frac{-b+-\sqrt{b^{2}-4ac}}{2a}$

To solve the quadratic equation by using the quadratic formula

1. Find the values of a , b and c from the equation

2. Substitute the values of a , b and c in the quadratic formula

3. Find the two values of x

∵ x² + 9x + 10 = 0

∴ a = 1 , b = 9 and c = 10

∵ $x=\frac{-9+\sqrt{(9)^{2}-4(1)(10)}}{2(1)}$

∴ $x=\frac{-9+\sqrt{81-40}}{2}$

∴ $x=\frac{-9+\sqrt{41}}{2}$

∴ x = -1.3

∵ $x=\frac{-9-\sqrt{(9)^{2}-4(1)(10)}}{2(1)}$

∴ $x=\frac{-9-\sqrt{81-40}}{2}$

∴ $x=\frac{-9-\sqrt{41}}{2}$

∴ x = -7.7

The solutions of the equations are -1.3 and -7.7

Learn more:

You can learn more about quadratic equation in brainly.com/question/8196933

#LearnwithBrainly

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Answer:

The Pearson's coefficient of correlation between the is 0.700.

Step-by-step explanation:

The correlation coefficient is a statistical degree that computes the strength of the linear relationship amid the relative movements of the two variables (i.e. dependent and independent).It ranges from -1 to +1.

The formula to compute correlation between two variables X and Y is:

$r(X, Y)=\frac{Cov(X, Y)}{\sqrt{V(X)\cdot V(Y)}}$

The formula to compute covariance is:

$Cov(X, Y)=n\cdot \sum XY-\sum X \cdot\sum Y$

The formula to compute the variances are:

$V(X)=n\cdot\sum X^{2}-(\sum X)^{2}\\V(Y)=n\cdot\sum Y^{2}-(\sum Y)^{2}$

Consider the table attached below.

Compute the covariance as follows:

$Cov(X, Y)=n\cdot \sum XY-\sum X \cdot\sum Y$

$=(5\times 165)-(30\times 25)\\=75$

Thus, the covariance is 75.

Compute the variance of X and Y as follows:

$V(X)=n\cdot\sum X^{2}-(\sum X)^{2}\\=(5\times 226)-(30)^{2}\\=230\\\\V(Y)=n\cdot\sum Y^{2}-(\sum Y)^{2}\\=(5\times 135)-(25)^{2}\\=50$

Compute the correlation coefficient as follows:

$r(X, Y)=\frac{Cov(X, Y)}{\sqrt{V(X)\cdot V(Y)}}$

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3 years ago

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A research firm tests the miles-per-gallon characteristics of three brands of gasoline. Because of different gasoline performanc

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Answer:

A.) At α = 0.05, is there a significant difference in the mean miles-per-gallon characteristics of the three brands of gasoline.

B.)The advantage of attempting to remove the block effect is

The completely randomized designs does not prove that H0 is incorrect only that it cannot be rejected.

Step-by-step explanation:

Using Two-way ANOVA method

Given problem

Observation I II III Row total (xr)

A 18 21 20 59

B 24 26 27 77

C 30 29 34 93

D 22 25 24 71

E 20 23 24 63

Col total (xc) 114 124 129 367

∑x²=9233→(A)

∑x²c/r

=1/5(114²+124²+129²)

=1/5(12996+15376+16641)

=1/5(45013)

=9002.6→(B)

∑x²r/c

=1/3(59²+77²+93²+71²+67²)

=1/3(3481+5929+8649+5041+4489)

=1/3(27589)

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(∑x)²/n

=(367)²/15

=134689/15

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SST=∑x²-(∑x)²/n

=(A)-(D)

=9233-8979.2667

=253.7333

Sum of squares between rows

SSR=∑x²r/c-(∑x)²/n

=(C)-(D)

=9196.3333-8979.2667

=217.0667

Sum of squares between columns

SSC=∑x²c/r-(∑x)²/n

=(B)-(D)

=9002.6-8979.2667

=23.3333

Sum of squares Error (residual)

SSE=SST-SSR-SSC

=253.7333-217.0667-23.3333

=13.3333

ANOVA table

Source Sums Degrees Mean Squares

of Variation of Squares of freedom

SS DF MS F p-value

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rows

B/w SSC=23.3333 c-1=2 MSC=11.6667 7 0.01

columns

Error (residual)SSE=13.3333 (r-1)(c-1)=8 MSE=1.6667

Total SST=253.7333 rc-1=14

Conclusion:

1. F for between Rows

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Therefore H0 is rejected

2. F for between Columns

The critical region for F(2,8) at 0.05 level of significance=4.459

We see that the calculated F for Colums=7>4.459

therefore H0 is rejected,and concluded that there is significant differentiating between columns

Part B:

To analyze the data for completely randomized designs click on anova two factor without replication in the data analysis dialog box of the excel spreadsheet.

The following table is obtained

Source DF Sum Mean F Statistic

(df1,df2) of Square (SS) Square (MS) P-value

Factor A 1 1496.5444 1496.5444 769.6514 0.001297

Rows

Factor B - 2 19.4444 9.7222 5 0.1667

Columns

Interaction

AB 2 3.8889 1.9444 0.1013 0.9045

Error 12 230.4 19.2

Total 17 1750.2778 102.9575

Factor - A- Rows

Since p-value < α, H0 is rejected.

Factor - B- Columns

Since p-value > α, H0 can not be rejected.

The averages of all groups assume to be equal.

Interaction AB

Since p-value > α, H0 can not be rejected.

The advantage of attempting to remove the block effect is

The completely randomized designs does not prove that H0 is incorrect only that it cannot be rejected.

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