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3 years ago

9

A man's salary is increased by $925 to $1036. Find the percentage increase. What would have been his new salary if the increase

had been 16%

1 answer:

exis [7]3 years ago

3 0

Answer:

See the solutions below

Step-by-step explanation:

Initial Salary= $925

Final salary= $1036

% increase= Change/initial *100

substitute

% increase= 1036-925/925 *100

% increase=111/925 *100

% increase=0.12 *100

% increase=12%

2. Let us find 16% of $925

=16/100*925

=0.16*925

=$148

Hence the salary would have been

=148+925

=$1073

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A 95% confidence interval for a population mean is computed from a sample of size 400. Another 95% confidence interval will be c

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Answer:

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Step-by-step explanation:

From the question we are told the confidence level is 95% , hence the level of significance is

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=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the 95% confidence interval is dependent on the value of the margin of error at a constant sample mean or sample proportion

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

Here assume that $Z_{\frac{\alpha }{2} } \ and \ \sigma \$ is constant so

$E = \frac{k}{\sqrt{n} }$

=> $E \sqrt{n} = K$

=> $E_1 \sqrt{n}_1 = E_2 \sqrt{n}_2$

So let $n_1 = 400$ and $n_2 = 100$

=> $E_1 \sqrt{400} = E_2 \sqrt{100}$

=> $E_1 = \frac{\sqrt{100} }{\sqrt{400} } E_2$

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So From this we see that the confidence interval for a sample size of 400 will be half that with a sample size of 100

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2 years ago

Find the following limit or state that it does not exist. ModifyingBelow lim With x right arrow minus 2 StartFraction 3 (2 x min

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Answer:

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Step-by-step explanation:

The objective is to state whether or not the following limit exists

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a_sh-v [17]

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3 years ago

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mixas84 [53]

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Vertex form

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So

vertex:-

(h,k)=(5,-9)

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