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1 year ago

1/2x=5/4 solve the equation and inequalities

Mathematics

1 answer:

aivan3 [116]1 year ago

3 0

The equation is simplified to x = 3/2

<h3>What are linear equations?</h3>

Linear equations are equations with the highest power of a variable as 1

Given the equation;

1/ 2x = 5/ 4

cross multiply

4 × x = 2 × 5

4x = 10

Make 'x' the subject

x = 10/ 4

Reduce to simplest term

x = 3/ 2

Thus, the equation is simplified to x = 3/2

Learn more about linear equations here:

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A random sample of 500 reports an average yearly income of $42,000 with a standard deviation of $1000. An estimate of the parame

Lorico [155]

Answer:

Step-by-step explanation:

Confidence Interval = mean + or - error margin

Mean = 42,000, error margin = width of estimate of the parameter ÷ 2 = 175 ÷ 2 = 87.50

We can be 95% confident that the population mean is 42,000 plus or minus 87.50

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3 years ago

on a map, point c is 4.3km due east of point b, while point b is 2.7km on a true bearing of 143 from a. find how far east point

topjm [15]

The correct answer is C

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3 years ago

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An article in The Engineer (Redesign for Suspect Wiring," June 1990) reported the results of an investigation into wiring errors

GarryVolchara [31]

Answer:

a) The 99% confidence interval on the proportion of aircraft that have such wiring errors is (0.0005, 0.0095).

b) A sample of 408 is required.

c) A sample of 20465 is required.

Step-by-step explanation:

Question a:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Of 1600 randomly selected aircraft, eight were found to have wiring errors that could display incorrect information to the flight crew.

This means that $n = 1600, \pi = \frac{8}{1600} = 0.005$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.005 - 2.575\sqrt{\frac{0.005*0.995}{1600}} = 0.0005$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.005 + 2.575\sqrt{\frac{0.005*0.995}{1600}} = 0.0095$

The 99% confidence interval on the proportion of aircraft that have such wiring errors is (0.0005, 0.0095).

b. Suppose we use the information in this example to provide a preliminary estimate of p. How large a sample would be required to produce an estimate of p that we are 99% confident differs from the true value by at most 0.009?

The margin of error is of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

A sample of n is required, and n is found for M = 0.009. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.009 = 2.575\sqrt{\frac{0.005*0.995}{n}}$

$0.009\sqrt{n} = 2.575\sqrt{0.005*0.995}$

$\sqrt{n} = \frac{2.575\sqrt{0.005*0.995}}{0.009}$

$(\sqrt{n})^2 = (\frac{2.575\sqrt{0.005*0.995}}{0.009})^2$

$n = 407.3$

Rounding up:

A sample of 408 is required.

c. Suppose we did not have a preliminary estimate of p. How large a sample would be required if we wanted to be at least 99% confident that the sample proportion differs from the true proportion by at most 0.009 regardless of the true value of p?

Since we have no estimate, we use $\pi = 0.5$

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.009 = 2.575\sqrt{\frac{0.5*0.5}{n}}$

$0.009\sqrt{n} = 2.575*0.5$

$\sqrt{n} = \frac{2.575*0.5}{0.009}$

$(\sqrt{n})^2 = (\frac{2.575*0.5}{0.009})^2$

$n = 20464.9$

Rounding up:

A sample of 20465 is required.

8 0

2 years ago

A rectangular prism has a length of 5 cm, a width of 7 cm, and a height of 4 cm. What is the volume of the prism?

Nataly [62]

Answer: 140

Step-by-step explanation:

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A student has $350.00 in her account and withdrew 3/5 of it . how much money , in dollars, remains in the account

Burka [1]

a whole, fraction wise, is always 1, so in this case we're looking at a whole of 5/5 = 1, and she withdrew 3/5, so what remained is 2/5

$\cfrac{5}{5}\qquad \qquad \cfrac{5}{5}-\cfrac{3}{5}\implies \cfrac{5-3}{5}\implies \cfrac{2}{5}~\hfill 350\left( \cfrac{2}{5} \right)\implies \boxed{140}$

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1 year ago