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3 years ago

(3•x)•3-2=1 I missed a lot of school and don’t know how to solve for x could you help me

Mathematics

1 answer:

elena-14-01-66 [18.8K]3 years ago

4 0

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$(3x) \times 3 - 2 = 1$

$(3x \times 3) - 2 = 1$

$9x - 2 = 1$

Add sides 2

$9x - 2 + 2 = 1 + 2$

Simplification

$9x = 3$

Divide sides by 9

$\frac{9x}{9} = \frac{3}{9} \\$

$\frac{9}{9} x = \frac{3}{9} \\$

$x = \frac{3}{3 \times 3} \\$

$x = \frac{1}{3} \\$

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Step-by-step explanation:

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3 years ago

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3 years ago

Suppose that the weights of passengers on a flight to Greenland on Frigid Aire Lines are normal with mean 175 pounds and standar

Natali5045456 [20]

Answer:

The minimum weight for a passenger who outweighs at least 90% of the other passengers is 203.16 pounds

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 175, \sigma = 22$

What is the minimum weight for a passenger who outweighs at least 90% of the other passengers?

90th percentile

The 90th percentile is X when Z has a pvalue of 0.9. So it is X when Z = 1.28. So

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 175}{22}$

$X - 175 = 22*1.28$

$X = 203.16$

The minimum weight for a passenger who outweighs at least 90% of the other passengers is 203.16 pounds

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3 years ago

Find the probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if:a) a

lara31 [8.8K]

Answer:

A) 0.0009765625

B) 0.0060466176

C) 2.7756 x 10^(-17)

Step-by-step explanation:

A) This problem follows a binomial distribution. The number of successes among a fixed number of trials is; n = 10

If a 0 bit and 1 bit are equally likely, then the probability to select in 1 bit is; p = 1/2 = 0.5

Now the definition of binomial probability is given by;

P(K = x) = C(n, k)•p^(k)•(1 - p)^(n - k)

Now, we want the definition of this probability at k = 10.

Thus;

P(x = 10) = C(10,10)•0.5^(10)•(1 - 0.5)^(10 - 10)

P(x = 10) = 0.0009765625

B) here we are given that p = 0.6 while n remains 10 and k = 10

Thus;

P(x = 10) = C(10,10)•0.6^(10)•(1 - 0.6)^(10 - 10)

P(x=10) = 0.0060466176

C) we are given that;

P((x_i) = 1) = 1/(2^(i))

Where i = 1,2,3.....,n

Now, the probability for the different bits is independent, so we can use multiplication rule for independent events which gives;

P(x = 10) = P((x_1) = 1)•P((x_2) = 1)•P((x_3) = 1)••P((x_4) = 1)•P((x_5) = 1)•P((x_6) = 1)•P((x_7) = 1)•P((x_8) = 1)•P((x_9) = 1)•P((x_10) = 1)

This gives;

P(x = 10) = [1/(2^(1))]•[1/(2^(2))]•[1/(2^(3))]•[1/(2^(4))]....•[1/(2^(10))]

This gives;

P(x = 10) = [1/(2^(55))]

P(x = 10) = 2.7756 x 10^(-17)

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3 years ago