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lukranit [14]
3 years ago
12

Y = 2x + 3 X Y -3 -1 2 4 5

Mathematics
2 answers:
antoniya [11.8K]3 years ago
4 0
Y = 2x + 3 \\  \\ x=-3 \\ Y = 2(-3) + 3=-6+3=-3\\  \\ x=-1 \\ Y = 2(-1) + 3=-2+3=1\\  \\ x=2 \\ Y = 2(2) + 3=4+3=7\\  \\ x=4 \\ Y = 2(4) + 3=8+3=11\\  \\ x=5 \\ Y = 2(5) + 3=10+3=13
Strike441 [17]3 years ago
3 0
Y=2x+3
for x = -3 y = 2*(-3)+3=-6+3=-3
for x = -1 y = 2*(-1)+3=-2+3=1
for x = 2 y = 2*2+3 = 4+3 = 7
for x = 4 y = 2*4+3 = 8+3 =11
for x = 5 y = 2*5+3 = 10+3 =13
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Data is collected to compare two different types of batteries. We measure the time to failure of 12 batteries of each type. Batt
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Using the t-distribution, it is found that since the <u>test statistic is greater than the critical value for the right-tailed test</u>, it is found that there is enough evidence to conclude that Battery B outlasts Battery A by more than 2 hours.

At the null hypothesis, it is <u>tested if it does not outlast by more than 2 hours</u>, that is, the subtraction is not more than 2:

H_0: \mu_B - mu_A \leq 2

At the alternative hypothesis, it is <u>tested if it outlasts by more than 2 hours</u>, that is:

H_1: \mu_B - \mu_A > 2

  • The sample means are: \mu_A = 8.65, \mu_B = 11.23
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Hence, the standard errors are:

s_{Ea} = S_{Eb} = \frac{0.67}{\sqrt{12}} = 0.1934

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s = \sqrt{s_{Ea}^2 + s_{Eb}^2} = \sqrt{0.1934^2 + 0.1934^2} = 0.2735

The test statistic is given by:

t = \frac{\overline{x} - \mu}{s}

In which \mu = 2 is the value tested at the hypothesis.

Hence:

t = \frac{\overline{x} - \mu}{s}

t = \frac{2.58 - 2}{0.2735}

t = 2.12

The critical value for a <u>right-tailed test</u>, as we are testing if the subtraction is greater than a value, with a <u>0.05 significance level</u> and 12 + 12 - 2 = <u>22 df</u> is given by t^{\ast} = 1.71

Since the <u>test statistic is greater than the critical value for the right-tailed test</u>, it is found that there is enough evidence to conclude that Battery B outlasts Battery A by more than 2 hours.

A similar problem is given at brainly.com/question/13873630

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