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3 years ago

Please help me solve step by step it's urgent

Mathematics

1 answer:

NNADVOKAT [17]3 years ago

6 0

Answer:

2i: 169.71

2ii: 0.17L

3a: 4×10⁻⁵

3b: 110011

Step-by-step explanation:

2i. The surface of the top and bottom of the tin is two times (top and bottom) π·r² = 2·π·3² = 18π cm².

The circumference of the circle is 2·π·r = 6π cm².

The area of the material connecting top and bottom is a rectangle of the tin height times the circumference: 6·6π = 36π cm².

This gives a total of 18π + 36π = 54π cm².

With π approximated by 22/7 the total surface area is 54*22/7 ≈ 169.71.

Notice how the calculation is simple by waiting until the very last moment to substitute π.

2ii. The volume is the area π·r² of the circle times the height of the tin: 9π*6 = 54π cm³ ≈ 169.71 cm³.

Since 1L = 1000 cm³ the volume is 0.16971 litres, which should be rounded to 0.17 L.

3a: If we rewrite P as 36 x 10⁻⁴ and realize that 36/2.25 = 16, then the fraction can be written as

16 x 10⁻⁴⁻⁶ = 16 x 10⁻¹⁰.

The square root of that is taking it to the power of 1/2, so (16x10⁻¹⁰)^0.5 = 4x10⁻⁵ = 0.00004

3b: 1111 1111 is 255 in decimal. 101 is 5 in decimal. 255/5 is 51 in decimal. 51 in binary is 110011.

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Step-by-step explanation:

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2 years ago

The circumference of a circle is 407 cm. Find the diameter, the radius, the length of an arc of 100°, and the area of a sector w

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4 0

3 years ago

An instructor in a college class recently gave an exam that was worth a total of 100 points. The instructor inadvertently made t

solniwko [45]

Answer:

B) Method 2 will increase the standard deviation of the students' scores.

Step-by-step explanation:

Given that an instructor in a college class recently gave an exam that was worth a total of 100 points.

The average score for his students was 43 and the standard deviation of the scores was 5 points.

And now he is considering two different strategies for rescaling the exam results of which:

Method 1 = Add 17 points to everyone's score.

Method 2 = Multiply everyone's score by 1.7 .

And we have to check what will be the impact of these methods on the standard deviation of the students' scores.

For this let us consider a simple example to understand this:

Firstly, Formula for calculating Standard Deviation = $\sqrt{\frac{\sum (X-Xbar)^{2}}{n-1}}$

Suppose,

X X - Xbar $(X - Xbar)^{2}$

3 3 - 6 = -3 -3 * -3 = 9

5 5 - 6 = -1 -1 * -1 = 1

10 10 - 6 = 4 4 * 4 = 16

Mean of above data, Xbar = $\frac{3+ 5+10}{3}$ = 6

Standard Deviation of data = $\sqrt{\frac{26}{3-1} }$ = 3.6055

Now let us suppose that we multiply each value of above data with 2 so the new data will be:

X X - Xbar $(X - Xbar)^{2}$

3*2 = 6 6 - 12 = -6 -6 * -6 = 36

5*2 = 10 10 - 12 = -2 -2 * -2 = 4

10*2 =20 20 - 12 = 8 8 * 8 = 64

Mean of new data, Xbar = $\frac{6+ 10+20}{3}$ = 12

Standard Deviation of new data = $\sqrt{\frac{104}{3-1} }$ = 7.2111

Hence, we see that when we multiply any value to the data the standard deviation will increase and in other words it will multiplied by that value which value we multiplied with each data value i.e. when we multiply each data value with 2 the standard deviation also get multiplied by as

3.6055 * 2 = 7.2111

Therefore option B is correct that Method 2 will increase the standard deviation of the students' scores.

And on the other hand Similarly by adding any constant to the data the Standard Deviation will remain same. Therefore Method 1 will have no impact on standard deviation of the students' scores.

8 0

3 years ago

Help this is due today!!!

kompoz [17]

Answer:

Hello!!

What is the value of m?

$0.61m-1.51m=9$