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DaniilM [7]
3 years ago
10

Solve for x. |2x - 8| > 6

Mathematics
2 answers:
AfilCa [17]3 years ago
6 0

Step-by-step explanation:

|2x-8|>6\iff 2x-8>6\ or\ 2x-86+8\ or\ 2x-8+814\ or\ 2x7\ or\ x

irina1246 [14]3 years ago
4 0

Answer:

Solution is x > 7 or x <  1.

Step-by-step explanation:

|2x - 8| > 6

This is equivalent to 2x - 8 > 6  or     2x - 8 < -6

2x - 8 > 6

2x > 14

x > 7.

2x - 8 < -6

2x < 2

x < 1.

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Bottles of grape soda are assumed to contain 300 milliliters of soda. There is some variation from bottle to bottle because the
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Because the p-value of 0.0828 is greater than the significance level of 0.05, we fail to reject the null hypothesis. We conclude the data provide convincing evidence that the mean amount of soda in all the bottles filled that day does not differ from the target value of 300 milliliters.

Step-by-step explanation:

The null hypothesis is that the sample mean does not differ from the population mean.

If p < α, we reject the null hypothesis.

If p > α, we do not reject the null hypothesis.

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In a certain population of the eastern thwump bird, the wingspan of the individual birds follows an approximately normal distrib
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a) P(48 < x < 58) = 0.576

b) P(X ≥ 1) = 0.9863

c) E(X) 2.88

d) P(x < 48) = 0.212

e) P(X > 2) = 0.06755

Step-by-step explanation:

The mean of the wingspan of the birds = μ = 53.0 mm

The standard deviation = σ = 6.25 mm

a) Probability of a bird having a wingspan between 48 mm and 58 mm can be found by modelling the problem as a normal distribution problem.

To solve this, we first normalize/standardize the two wingspans concerned.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ

For wingspan 48 mm

z = (48 - 53)/6.25 = - 0.80

For wingspan 58 mm

z = (58 - 53)/6.25 = 0.80

To determine the probability that the wingspan of the first bird chosen is between 48 and 58 mm long. P(48 < x < 58) = P(-0.80 < z < 0.80)

We'll use data from the normal probability table for these probabilities

P(48 < x < 58) = P(-0.80 < z < 0.80) = P(z < 0.8) - P(z < -0.8) = 0.788 - 0.212 = 0.576

b) The probability that at least one of the five birds has a wingspan between 48 and 58 mm = 1 - (Probability that none of the five birds has a wingspan between 48 and 58 mm)

P(X ≥ 1) = 1 - P(X=0)

Probability that none of the five birds have a wingspan between 48 and 58 mm is a binomial distribution problem.

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = number of birds = 5

x = Number of successes required = number of birds with wingspan between 48 mm and 58 mm = 0

p = probability of success = Probability of one bird having wingspan between 48 mm and 58 mm = 0.576

q = probability of failure = Probability of one bird not having wingspan between 48 mm and 58 mm = 1 - 0.576 = 0.424

P(X=0) = ⁵C₀ (0.576)⁰ (1 - 0.576)⁵ = (1) (1) (0.424)⁵ = 0.0137

The probability that at least one of the five birds has a wingspan between 48 and 58 mm = P(X≥1) = 1 - P(X=0) = 1 - 0.0137 = 0.9863

c) The expected number of birds in this sample whose wingspan is between 48 and 58 mm.

Expected value is a sum of each variable and its probability,

E(X) = mean = np = 5×0.576 = 2.88

d) The probability that the wingspan of a randomly chosen bird is less than 48 mm long

Using the normal distribution tables again

P(x < 48) = P(z < -0.8) = 1 - P(z ≥ -0.8) = 1 - P(z ≤ 0.8) = 1 - 0.788 = 0.212

e) The probability that more than two of the five birds have wingspans less than 48 mm long = P(X > 2) = P(X=3) + P(X=4) + P(X=5)

This is also a binomial distribution problem,

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = number of birds = 5

x = Number of successes required = number of birds with wingspan less than 48 mm = more than 2 i.e. 3,4 and 5.

p = probability of success = Probability of one bird having wingspan less than 48 mm = 0.212

q = probability of failure = Probability of one bird not having wingspan less than 48 mm = 1 - p = 0.788

P(X > 2) = P(X=3) + P(X=4) + P(X=5)

P(X > 2) = 0.05916433913 + 0.00795865476 + 0.00042823218

P(X > 2) = 0.06755122607 = 0.06755

5 0
3 years ago
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