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Ilya [14]
2 years ago
13

Please look at picture

Mathematics
1 answer:
Alex_Xolod [135]2 years ago
6 0
6x - 7 = 4x + 7
2x = 14
x = 7

The answer is 7
You might be interested in
Write -4 1/8 as a decimal
Hatshy [7]
-4 1/8 as decimal.
The decimal is -3.875.
7 0
3 years ago
If 3t - 7 = 5t, then 6t = <br> A. 21 <br> B. -7 <br> C. -21<br> D. -42
podryga [215]

Answer:

C. -21 is your answer

Step-by-step explanation:

Solve for t. Isolate the variable t in the first equation, then use the number gotten to solve the second equation.

3t - 7 = 5t

First, subtract 3t from both sides

3t (-3t) - 7 = 5t (-3t)

-7 = 5t - 3t

-7 = 2t

Isolate the variable (t). Divide 2 from both sides

(-7)/2 = (2t)/2

t = -7/2

t = -3.5

Plug in -3.5 for t in the second equation

6(t) =

6(-3.5) =

6(-3.5) = -21

C. -21 is your answer

~

6 0
2 years ago
1)A System of equations is shown below . What is the solution to the system of equations? 5x+2y=-15 2x-2y=-6
meriva

Answer:

x= -3 and y= 0

Step-by-step explanation:

5x+2y=-15

<u>2x-2y=-6     </u>

<u>7x        =-21</u>

x= -3

Putting value of x in equation 1  

5(-3) +2y=-15

-15+2y= -15

2y= 0

y= 0

This can be solved with the help of matrices

In matrix form the above equations can be written in the form

\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]  \left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]

Let

\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right] = A  \left[\begin{array}{ccc}x\\y\\\end{array}\right]  = X  and  \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]= B

Then AX= B

or X= A⁻¹ B

where  A⁻¹= adj A/ ║A║   where mod A≠ 0

adj A=  \left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]

║A║= ( 5*-2- 2*2)= -10-4= -14≠0

X= A⁻¹ B

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]    =- 1/14  \left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]   \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]    =- 1/14     \left[\begin{array}{ccc}-2*-15&+ -2*-6\\-2*-15&+ 5*-6\\\end{array}\right]

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]  =- 1/14 \left[\begin{array}{ccc} 30&+12\\30&+-30\\\end{array}\right]

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]  =- 1/14 \left[\begin{array}{ccc}42\\0\\\end{array}\right]

\left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-42/14\\0/-14\\\end{array}\right]

\left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-3\\0\\\end{array}\right]

From here x= -3 and y= 0

Solution Set = [(-3,0)]

3 0
3 years ago
A national grocery store chain wants to test the difference in the average weight of turkeys sold in Detroit and the average wei
Arte-miy333 [17]

Answer:

<em>Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom</em>

<em>The null hypothesis is accepted . </em>

<em>Assume the population variances are approximately the same</em>

<u><em>Step-by-step explanation:</em></u>

<u>Explanation</u>:-

Given data a random sample of 20 turkeys sold at the chain's stores in Detroit yielded a sample mean of 17.53 pounds, with a sample standard deviation of 3.2 pounds

<em>The first sample size  'n₁'= 20</em>

<em>mean of the first sample 'x₁⁻'= 17.53 pounds</em>

<em>standard deviation of first sample  S₁ = 3.2 pounds</em>

Given data a random sample of 24 turkeys sold at the chain's stores in Charlotte yielded a sample mean of 14.89 pounds, with a sample standard deviation of 2.7 pounds

<em>The second sample size  n₂ = 24</em>

<em>mean of the second sample  "x₂⁻"= 14.89 pounds</em>

<em>standard deviation of second sample  S₂ =  2.7 poun</em>ds

<u><em>Null hypothesis</em></u><u>:-</u><u><em>H₀</em></u><em>: The Population Variance are approximately same</em>

<u><em>Alternatively hypothesis</em></u><em>: H₁:The Population Variance are approximately same</em>

<em>Level of significance ∝ =0.05</em>

<em>Degrees of freedom ν = n₁ +n₂ -2 =20+24-2 = 42</em>

<em>Test statistic :-</em>

<em>    </em>t = \frac{x^{-} _{1} -  x_{2} }{\sqrt{S^2(\frac{1}{n_{1} } }+\frac{1}{n_{2} }  }

<em>    where         </em>S^{2}   = \frac{n_{1} S_{1} ^{2}+n_{2}S_{2} ^{2}   }{n_{1} +n_{2} -2}

                      S^{2} = \frac{20X(3.2)^2+24X(2.7)^2}{20+24-2}

<em>              substitute values and we get  S² =  40.988</em>

<em>     </em>t= \frac{17.53-14.89 }{\sqrt{40.988(\frac{1}{20} }+\frac{1}{24}  )}<em></em>

<em>  </em>   t =  1.3622

  Calculated value t = 1.3622

Tabulated value 't' =  2.081

Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom

<u><em>Conclusion</em></u>:-

<em>The null hypothesis is accepted </em>

<em>Assume the population variances are approximately the same.</em>

<em>      </em>

<em>                        </em>

<em>                    </em>

6 0
2 years ago
What is the area of the parallelogram
GaryK [48]
Easy:

formula for parallelogram is b times h

so 5 times 14 =

70 sq mm.
3 0
2 years ago
Read 2 more answers
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