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Umnica [9.8K]
3 years ago
9

Which value satisfies the inequality -2x + 8 + 5x > 2x + 1? A) -15 B) -10 C) -7 D) -5

Mathematics
2 answers:
larisa86 [58]3 years ago
8 0

It’s D.) -5

If you solve it you will know x>-7 and the only value that satisfies the equality is -5.

Novosadov [1.4K]3 years ago
5 0
<span>-2x + 8 + 5x > 2x + 1
3x + 8 > 2x + 1
         x > -7

answer
D) - 5</span>
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You work for a soft-drink company in the quality control division. You are interested in the standard deviation of one of your p
frutty [35]

Answer:

Step-by-step explanation:

weight (x)                         ( x - \overline x)^2

12.05                               0.00123

12                                     0.00019

11.96                                0.00292

12.14                                0.0159

12.12                                0.00112

11.97                                0.0019

11.96                               0.0029

12.11                                 0.0092

11.94                                 0.0055

11.82                                0.038

12                                     0.00019

12.18                                0.028

11.91                                  0.0108

12.06                                0.0021

11.99                                  0.0029

\sum x = 180.21                    \sum (x - \overline x)^2 = 0.12285

Sample size = 15

\overline x = \dfrac{180.21}{15} = 12.014

Sample standard deviation

x = \sqrt{\dfrac{0.12285}{15-1}}

x = 0.0937

\text{degree of freedom = n - 1}

degree of freedom = 15- 1=14

\text{confidence interval} \alpha = 1- 0.98 = 0.02 \\ \\ \alpha/2 = 0.02/2 = 0.01

From \  X^2  \  table}\text{, the critical value} X^2 \text{=0.99 for degree of freedom =14 is given by : 4.660}

\text{, the critical value} X^2 \text{=0.01 for degree of freedom =14 is given by : 29.141}

\text{Thus, the lower limit = 4.660, the upper limit = 29.141}

\text{the confidence interval of the standard deviation} \ \sigma \  is:

\sqrt{\dfrac{(n -1)s^2}{X^2_{\dfrac{\sigma}{2}}} } \le \sigma \le \sqrt{\dfrac{(n -1)s^2}{X^2_{1-\dfrac{\sigma}{2}}} }

replacing our values:

\sqrt{\dfrac{(15 -1)0.0937^2}{29.141} } < \sigma

= 0.06495 < \sigma < 0.1624

\mathbf{Thus; \  \sigma (0.06495 , 0.1624)}

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