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3 years ago

How to find inequalities

Mathematics

1 answer:

geniusboy [140]3 years ago

6 0

Answer:

Step 1 Eliminate fractions by multiplying all terms by the least common denominator of all fractions.

Step 2 Simplify by combining like terms on each side of the inequality.

Step 3 Add or subtract quantities to obtain the unknown on one side and the numbers on the other.

Hope that helped happy holidays!

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The American Management Association is studying the income of store managers in the retail industry. A random sample of 49 manag

VashaNatasha [74]

Answer:

a) The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.

b) The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.

Step-by-step explanation:

Question a:

We have to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a p-value of $1 - 0.025 = 0.975$ , so Z = 1.96.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96\frac{2050}{\sqrt{49}} = 574$

The lower end of the interval is the sample mean subtracted by M. So it is 45420 - 574 = $44,846.

The upper end of the interval is the sample mean added to M. So it is 45420 + 574 = $45,994.

The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.

Question b:

The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.

5 0

3 years ago

Sarah is 50 years old. she is twice as old as her friend bill and bill is 5 times as old as his sister jane. jane's mom is 12 ti

Pavlova-9 [17]

10 years. Bill is 25, his sister is 5, and their mother is 60.

7 0

3 years ago

Read 2 more answers

The table shows a student's proof of the quotient rule for logarithms.

nikklg [1K]

Answer:

The error is at step (3) .

The correct step (3) will be,

$\log_{b}(\frac {b^{x}}{b^{y}})$

= $\log_{b}(b^{x - y})$ [by using the laws of indices]

All other steps are correct.

Step-by-step explanation:

The error is at the step (3) , because the student has tried to prove the quotient rule of logarithms by using the property i.e., 'The quotient rule of logarithm' itself , i.e. ,by assuming the property does hold before proving it. So, the proof is fallacious.

The correct step (3) will be,

$\log_{b}(\frac {b^{x}}{b^{y}})$

= $\log_{b}(b^{x - y})$ [by using the laws of indices]

All other steps are correct.

5 0

3 years ago

The Town of Hertfordshire clerk knows that 23% of dogs in the town have completed emotional support training. Hertfordshire plan

Nataly_w [17]

Answer:

95.64% probability that under 30% of the dogs are emotional support trained

Step-by-step explanation:

For each dog, there are only two possible outcomes. Either they have completed emotional support training, or they have not. So we use the binomial probability distribution to solve this problem.

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .