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3 years ago

I need much much much much help with my Alogrithm skills

Mathematics

1 answer:

Pie3 years ago

3 0

2×8=16 bring down the 6 carry the one 2×1=2+1=3 so you have 36
bring down a 0 3×8=24 carry the 2 bring down the 4 3×1=3+2=5 which is 540
540+36=576
the answer is 576

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Which value is equivalent to 3/11? OA A. 0.27 O B. 0.27 OC C. 3.6 OD. 3.

Sladkaya [172]

Answer:

0.27 repeating.

Step-by-step explanation:

So to solve this we just have to use division.

1 clearly doesnt go into 3 at least one time, so we can add a decimal point and add a 0 to make it 30.

11 goes into 30 2 times, so we have:

0.2

and 30-22=8

So we can add another 0 and make it 80.

Then 11 goes into 80 7 times. So we have:

0.27

and 80-77=3

So again, add the 0, we have 30.

11 goes into 30 2 times, so:

0.272

and 30-22=8

Add another 0 we get 80.

11 goes into 80 7 times.

So finally, we have:

0.2727.

This is a repeating decimal.

This can be shown as:

0.27

So this is your answer!

I was a bit confused with which one it was on your answer key, but knowing that it is 0.27 I am guessing you can chose!

Hope this helps!

4 0

3 years ago

If Tom has 3 different sweaters, and 5 different pairs of pants, how many different combinations could he wear?

Vladimir [108]

Check the tree diagram in the picture.

Tom could pick Pants 1 (P1), then Sweater 1, 2 or 3. So 3 combinations

Similarly if we picked P2, P3, P4 or P5 we would have 3 combinations per each.

In total we have 5*3=15 different combinations.

Answer:15

7 0

3 years ago

Write an equation of a line that passes through the point (1, 2) and is perpendicular to the line y=-1/4x+2

Dmitriy789 [7]

Answer:

y = 4x - 2 would be the equation

7 0

3 years ago

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 $\pi = 0.5$