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Leokris [45]
3 years ago
14

3x-2=25 solve equation

Mathematics
2 answers:
ki77a [65]3 years ago
6 0

Answer:

x=9

Step-by-step explanation:

3x-2=25

Add 2 to each side

3x-2+2=25+2

3x = 27

Divide by 3

3x/3 = 27/3

x = 9

VARVARA [1.3K]3 years ago
6 0

Answer:

x=9

Step-by-step explanation:

3x-2=25

3x =25+2

3x =27

x =27/3

x =9

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Shelia's measured glucose level one hour after a sugary drink varies according to the normal distribution with μ = 117 mg/dl and
TiliK225 [7]

Answer:

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

Step-by-step explanation:

To solve this problem, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

Central limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 117, \sigma = 10.6, n = 6, s = \frac{10.6}{\sqrt{6}} = 4.33

What is the level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L ?

This is the value of X when Z has a pvalue of 1-0.01 = 0.99. So X when Z = 2.33.

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

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

2.33 = \frac{X - 117}{4.33}

X - 117 = 2.33*4.33

X = 127.1

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

6 0
3 years ago
Find a third-degree polynomial equation with rational coefficients that has roots -4 and 6 + i.
dolphi86 [110]

Answer:

x³ - 8x² - 11x + 148

Step-by-step explanation:

Given that x = 6 + i is a root then x = 6 - i is also a root

Complex roots occur as conjugate pairs.

The factors are therefore (x - (6 + i)) and(x - (6 - i))

Given x = - 4 is a root then (x + 4) is a factor

The polynomial is the product of the factors, that is

p(x) = (x + 4)(x - (6 + i))(x - (6 - i))

      = (x + 4)(x - 6 - i)(x - 6 + i)

      = (x + 4)((x - 6)² - i²)

      = (x + 4)(x² - 12x + 36 + 1)

      = (x + 4)(x² - 12x + 37) ← distribute

      = x³ + 4x² - 12x² - 48x + 37x + 148

      = x³ - 8x² - 11x + 148

 

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3 years ago
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antoniya [11.8K]

Answer:

Not sure

Step-by-step explanation:

Sorry

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