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lianna [129]
3 years ago
6

6+x3=1

Mathematics
1 answer:
Vadim26 [7]3 years ago
6 0

Answer:

6 +  \frac{x}{3}  = 1 \\  \frac{x}{3}  = 1 - 6 \\  \frac{x}{3}  =  - 5 \\ x = 3 \times ( - 5) \\  \boxed{x =  - 15}✓

★ 6+x/3=1

What is the solution to the equation?

  • x=−21
  • x=−5
  • x = 21
  • <u>x=−15</u>
<h3><u>x= -15</u> is the right answer.</h3>
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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

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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

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               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.

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<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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