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melamori03 [73]
3 years ago
15

4x - 2y = -20 7x + 2y = -2

Mathematics
1 answer:
Brut [27]3 years ago
8 0

Answer:

(-2, 6)

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Order of Operations: BPEMDAS
  • Equality Properties

<u>Algebra I</u>

  • Solving systems of equations using substitution/elimination
  • Solving systems of equations by graphing

Step-by-step explanation:

<u>Step 1: Define systems</u>

4x - 2y = -20

7x + 2y = -2

<u>Step 2: Rewrite systems</u>

4x - 2y = -20

  1. Add 2y to both sides:                    4x = 2y - 20
  2. Divide 4 on both sides:                 x = 1/2y - 5

<u>Step 3: Redefine systems</u>

x = 1/2y - 5

7x + 2y = -2

<u>Step 4: Solve for </u><em><u>y</u></em>

<em>Substitution</em>

  1. Substitute in <em>x</em>:                         7(1/2y - 5) + 2y = =-2
  2. Distribute 7:                              7/2y - 35 + 2y = -2
  3. Combine like terms:                11/2y - 35 = -2
  4. Add 35 to both sides:              11/2y = 33
  5. Isolate <em>y</em>:                                   y = 6

<u>Step 5: Solve for </u><em><u>x</u></em>

  1. Define original equation:                    7x + 2y = -2
  2. Substitute in <em>y</em>:                                     7x + 2(6) = -2
  3. Multiply:                                                7x + 12 = -2
  4. Subtract 12 on both sides:                   7x = -14
  5. Divide 7 on both sides:                        x = -2

<u>Step 6: Graph systems</u>

<em>Check the system.</em>

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3 years ago
An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 37 type K batteries and a sample of 58
Olegator [25]

Answer:

Hypothesis Test states that we will accept null hypothesis.

Step-by-step explanation:

We are given that an engineer is comparing voltages for two types of batteries (K and Q).

where, \mu_1 = true mean voltage for type K batteries.

           \mu_2 = true mean voltage for type Q batteries.

So, Null Hypothesis, H_0 :  \mu_1 = \mu_2 {mean voltage for these two types of

                                                        batteries is same}

Alternate Hypothesis, H_1 : \mu_1 \neq \mu_2 {mean voltage for these two types of

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<em>The test statistics we use here will be :</em>

                     \frac{(X_1bar-X_2bar) - (\mu_1 - \mu_2) }{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } }  follows t_n__1 + n_2  -2

where, X_1bar = 8.54         and     X_2bar = 8.69

                s_1  = 0.225       and         s_2     =  0.725

               n_1   =  37           and         n_2     =  58

               s_p = \sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2}  }{n_1+n_2-2} } =   \sqrt{\frac{(37-1)0.225^{2}+(58-1)0.725^{2}  }{37+58-2} }  =  0.585               Here, we use t test statistics because we know nothing about population standard deviations.

     Test statistics =  \frac{(8.54-8.69) - 0 }{0.585\sqrt{\frac{1}{37}+\frac{1}{58}  } } follows t_9_3

                             = -1.219

<em>At 0.1 or 10% level of significance t table gives a critical value between (-1.671,-1.658) to (1.671,1.658) at 93 degree of freedom. Since our test statistics is more than the critical table value of t as -1.219 > (-1.671,-1.658) so we have insufficient evidence to reject null hypothesis.</em>

Therefore, we conclude that mean voltage for these two types of batteries is same.

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Step-by-step explanation:

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mafiozo [28]
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Hope This Helps You!
<span>Good Luck Studying :)</span>
7 0
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