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pashok25 [27]
2 years ago
13

Solve for y

Mathematics
2 answers:
Andreas93 [3]2 years ago
8 0

Answer:

y = 3

Step-by-step explanation:

2y - 3 = \sqrt{3y^2-10y+12}

square both sides to remove sqrt bracket

(2y - 3)^2 = ( \sqrt{3y^2-10y+12} )^2

simplify both sides

(2y - 3)(2y - 3) = 3y^2 - 10y + 12

4y^2 - 12y + 9 = 3y^2 - 10y + 12

bring all value to left side

y^2 - 2y - 3 = 0

factor

(y - 3)(y + 1)

solve for y

y = 3, y = -1

When plugged back into the equation, only y = 3 is true

Firlakuza [10]2 years ago
4 0

Answer:

y = 3

Step-by-step explanation :

2y - 3 = \sqrt{ 3y² - 10y + 12}

Swap the sides both of the equation.

\sqrt{ 3y² - 10y + 12} = 2y - 3

To remove the brackets of equations square both side and simplify .

3y² - 10y + 12 = 4y² - 12y + 9

Move the expression to left-hand side and change its sign.

3y² - 10y + 12 - 4y² + 12y - 9 = 0

collect like terms

3y² - 4y² - 10y + 12y + 12 - 9 = 0

-y² + 2y + 3 = 0

Change the sign of expression. because it helps to solve.

y² - 2y - 3 = 0

Splits the term -2y

<u>y² + y</u> <u>-</u><u>3y - 3</u> = 0

Factor out y from the first pair and -3 from second pair of expression.

y ( y + 1 ) - 3 ( y + 1) = 0

Factor out y + 1 from the expression.

( y + 1 ) ( y - 3 ). = 0

When product and factors equals 0. at least one factor is 0.

y + 1 =0

y - 3 = 0

Solve for y

y = -1 and y = 3

If we plug the 3 as y in the expression we find that y = 3 is the true solution of this expression.

This equation has one solution which is y = 3.

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maxonik [38]

Answer:

z=\frac{5.0611-5}{\frac{0.2803}{\sqrt{42}}}=1.413    

p_v =2*P(z>1.413)=0.158  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance

Step-by-step explanation:

Data given and notation  

\bar X=5.0611 represent the sample mean

\sigma=0.2803 represent the population standard deviation for the sample  

n=42 sample size  

\mu_o =5 represent the value that we want to test

\alpha=0.1 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 5, the system of hypothesis would be:  

Null hypothesis:\mu = 5  

Alternative hypothesis:\mu \neq 5  

If we analyze the size for the sample is > 30 but and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

z=\frac{5.0611-5}{\frac{0.2803}{\sqrt{42}}}=1.413    

P-value

Since is a two sided test the p value would be:  

p_v =2*P(z>1.413)=0.158  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance

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Answer:

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