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iren2701 [21]
3 years ago
10

Evaluate 8(7x--3) when x=9

Mathematics
1 answer:
Schach [20]3 years ago
5 0

I will assume that by "Evaluate 8(7x--3) when x=9" you meant:

"Evaluate 8(7x-3) when x=9."

First, substitute 9 for x in 7x-3: 7(9)-3 = 60. Rewrite the problem as:

8(60). Multiply. Answer: 480

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neonofarm [45]

Answer:

2/5 or .4

Step-by-step explanation:

basically look at the points where the line meets the graph. two points would be 0,20 and 50,40. then count the amount that it changes vertically and then horizontally. +20vertical +50 horizonal giving the slope 20/50. this can then be simplified to 2/5 or the equivalent number .4

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GarryVolchara [31]

Answer:

\mu_p -\sigma_p = 0.74-0.0219=0.718

\mu_p +\sigma_p = 0.74+0.0219=0.762

68% of the rates are expected to be betwen 0.718 and 0.762

\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696

\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784

95% of the rates are expected to be betwen 0.696 and 0.784

\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674

\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806

99.7% of the rates are expected to be betwen 0.674 and 0.806

Step-by-step explanation:

Check for conditions

For this case in order to use the normal distribution for this case or the 68-95-99.7% rule we need to satisfy 3 conditions:

a) Independence : we assume that the random sample of 400 students each student is independent from the other.

b) 10% condition: We assume that the sample size on this case 400 is less than 10% of the real population size.

c) np= 400*0.74= 296>10

n(1-p) = 400*(1-0.74)=104>10

So then we have all the conditions satisfied.

Solution to the problem

For this case we know that the distribution for the population proportion is given by:

p \sim N(p, \sqrt{\frac{p(1-p)}{n}})

So then:

\mu_p = 0.74

\sigma_p =\sqrt{\frac{0.74(1-0.74)}{400}}=0.0219

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

\mu_p -\sigma_p = 0.74-0.0219=0.718

\mu_p +\sigma_p = 0.74+0.0219=0.762

68% of the rates are expected to be betwen 0.718 and 0.762

\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696

\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784

95% of the rates are expected to be betwen 0.696 and 0.784

\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674

\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806

99.7% of the rates are expected to be betwen 0.674 and 0.806

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Likurg_2 [28]

X x 4 *equal sign* 64 hope this helps. :)

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The answer would be B

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