PLEASE HELP WITH THE QUESTION BELOW!!!

Can someone show me how to work out this example? (5+44+182−20−10)÷(2+5)

lora16 [44]

Answer:

approximately 35.9

Step-by-step explanation:

(5+44+182-20-10)÷(2+5)

(99+182-20-10)÷(7)

(281-20-10)÷(7)

(261-10)÷÷(7)

(251)÷(7)

approximately 35.9

6 0

4 years ago

Please help this is important

Galina-37 [17]

Answer:

First blank - SSS congruent condition

Second blank - Congruent 

5 0

3 years ago

Read 2 more answers

Someone help me please

galben [10]

Answer:

∛27 = 3

Step-by-step explanation:

A radical is simply a fractional exponent: $a^{(\frac{m}{n})} = \sqrt[n]{a^{m} }$

Hence, ∛27 = $27^{(\frac{1}{3})}$

Since 27 = 3³, then:

You could rewrite ∛27 as ∛(3)³.

$\sqrt[3]{3^{(3)} } = 3^{[(3)*(\frac{1}{3})]}$

Multiplying the fractional exponents (3 × 1/3) will result in 1 (because 3 is the multiplicative inverse of 1/3). The multiplicative inverse of a number is defined as a number which when multiplied by the original number gives the product as 1.

Therefore, ∛27 = 3.

3 0

3 years ago

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If 10 is to the second power then divided by 3 and multiplied by 7 what’s the answer?

Savatey [412]

First, 10 to the second power equals to 100. 100 divided by 3 equals to 33.33. 33.33 x 7 equals to 233.33. So the answer is 233.33.

4 0

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 $\pi = 0.5$