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lutik1710 [3]
3 years ago
7

Let x be a positive number such that the distance between x and -3.8 on a number line is exactly 2 times the distance between x

and 1.7. What is the value of x? Express your answer as a decimal to the nearest tenth.
I know the answer is 7.2 but idk why. HELP PLZZ
Mathematics
1 answer:
anyanavicka [17]3 years ago
3 0

Answer:

x = 7.2

Step-by-step explanation:

Let x be a positive number such that the distance between x and -3.8 on a number line is exactly 2 times the distance between x and 1.7.

Now. the distance between - 3.8 and x will be (x + 3.8) and the distance between x and 1.7 will be (x - 1.7) {Since x is a positive number}

(Assuming x is greater than 1.7}

Given that, x + 3.8 = 2(x - 1.7)

⇒ x + 3.8 = 2x - 3.4

⇒ x = 7.2 (Answer)

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Step-by-step explanation:

In statistics, the empirical rule states that for a normally distributed random variable,

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In mathematical notation, as shown in the figure below (for a standard normal distribution), the empirical rule is described as

                             \Phi(\mu \ - \ \sigma \ \leq X \ \leq \mu \ + \ \sigma) \ = \ 0.6827 \qquad (4 \ \text{s.f.}) \\ \\ \\ \Phi(\mu \ - \ 2\sigma \ \leq X \ \leq \mu \ + \ 2\sigma) \ = \ 0.9545 \qquad (4 \ \text{s.f.}) \\ \\ \\ \Phi}(\mu \ - \ 3\sigma \ \leq X \ \leq \mu \ + \ 3\sigma) \ = \ 0.9973 \qquad (4 \ \text{s.f.})

where the symbol \Phi (the uppercase greek alphabet phi) is the cumulative density function of the normal distribution, \mu is the mean and \sigma is the standard deviation of the normal distribution defined as N(\mu, \ \sigma).

According to the empirical rule stated above, the interval that contains the prices of 99.7% of college textbooks for a normal distribution N(113, \ 12),

                \Phi(113 \ - \ 3 \ \times \ 12 \ \leq \ X \ \leq \ 113 \ + \ 3 \ \times \ 12) \ = \ 0.9973 \\ \\ \\ \-\hspace{1.75cm} \Phi(113 \ - \ 36 \ \leq \ X \ \leq \ 113 \ + \ 36) \ = \ 0.9973 \\ \\ \\ \-\hspace{3.95cm} \Phi(77 \ \leq \ X \ \leq \ 149) \ = \ 0.9973

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