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Lina20 [59]
2 years ago
12

Please help giving Brainiest

Mathematics
2 answers:
Anna [14]2 years ago
7 0

(1/2)²-6(2-2/3)

(1/2)²-6(2-2/3)

LCM=6

1/2²-6(2/1-2/3)

12-4

______

6

1/2²-6(8/6)

1-(48/36)

1-(4/3)

= -4/3

Elanso [62]2 years ago
4 0

Answer:

Correct answer is - 31/4

Step-by-step explanation:

Have a good day :)

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3 years ago
A consumer group has determined that the distribution of life spans for gas ovens has a mean of 15.0 years and a standard deviat
Mnenie [13.5K]

Answer:

B. Mean = 1.6 years, standard deviation = 0.92 years, shape: approximately Normal.

Step-by-step explanation:

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

Normal Probability Distribution:

Problems of normal distributions can be 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 normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Subtraction of normal variables:

When we subtract normal variables, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

35 gas ovens

A consumer group has determined that the distribution of life spans for gas ovens has a mean of 15.0 years and a standard deviation of 4.2 years. This means that:

\mu_G = 15, \sigma_G = 4.2, n = 35, s_G = \frac{4.2}{\sqrt{35}} = 0.71

40 electric ovens.

The distribution of life spans for electric ovens has a mean of 13.4 years and a standard deviation of 3.7 years.

\mu_E = 13.4, \sigma_E = 3.7, n = 40, s_E = \frac{3.7}{\sqrt{40}} = 0.585

Which of the following best describes the sampling distribution of barXG - bar XE, the difference in mean life span of gas and electric ovens?

By the Central Limit Theorem, the shape is approximately normal.

Mean: \mu = \mu_G - \mu_E = 15 - 13.4 = 1.6

Standard deviation:

s = \sqrt{s_G^2+s_E^2} = \sqrt{(0.71)^2+(0.585)^2} = 0.92

So the correct answer is given by option b.

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Step-by-step explanation:

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