What is the Volume a cube with a edge length of 7.3 cm

Evaluate the principal square root of 64

anygoal [31]

The square root of 64 is 8 because 8x8 is 64

3 0

3 years ago

The calibration of a scale is to be checked by weighing a 13 kg test specimen 25 times. Suppose that the results of different we

Natali5045456 [20]

Solution :

a).

Given : Number of times, n = 25

Sigma, σ = 0.200 kg

Weight, μ = 13 kg

Therefore the hypothesis should be tested are :

$H_0 : \mu = 13$

$H_a : \mu \neq 13$

b). When the value of $\overline x = 12.84$

Test statics :

$Z=\frac{(\overline x - \mu)}{\frac{\sigma}{\sqrt n}}$

$Z=\frac{(14.82-13)}{\frac{0.2}{\sqrt {25}}}$

$=\frac{1.82}{0.04}$

= 45.5

P-value = 2 x P(Z > 45.5)

= 2 x 1 -P (Z < 45.5) = 0

Reject the null hypothesis if P value < α = 0.01 level of significance.

So reject the null hypothesis.

Therefore, we conclude that the true mean measured weight differs from 13 kg.

3 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

What is the interquartile range of the data set

Verizon [17]

There are many methods of doing this. This one finds the median; however, instead of halving, you can use quarters instead.

First order the data, ascendingly:

10, 17, 18, 18, 18, 20, 20, 25, 25, 30, 30

Find the median:

For an even number of values: The number of values / 2

For an odd number of values: The number of values + 1 / 2

11 is odd.

11 + 1 / 2

12 / 2 = 6

The median is at the 6th position.

10, 17, 18, 18, 18, 20, 20, 25, 25, 30, 30

The median is 20.

Other method:

You don't need this step.

Using the values lower than the median, exclusive, find the median of the lower values;

For an odd number of values: The number of values + 1 / 2

10, 17, 18, 18, 18

5 + 1 / 2 = 6 / 2 = 3

The median of the lower values is at position 3.

This is the lower quartile.

10, 17, 18, 18, 18

Other method:

For an odd number of values: The number of values + 1 / 4

11 + 1 / 4 = 12 / 4 = 3

10, 17, 18, 18, 18, 20, 20, 25, 25, 30, 30

Using the values greater than the median, exclusive, find the median:

For an odd number of values: The number of values + 1 / 2

20, 25, 25, 30, 30

5 + 1 / 2 = 6 / 2 = 3

The median of the upper values is at position 3.

20, 25, 25, 30, 30

This is 25.

This is the upper quartile.

Other method:

For an odd number of values: (The number of values + 1 / 4) * 3

(11 + 1 / 4) * 3 = (12 / 4) * 3 = 3 * 3 = 9

10, 17, 18, 18, 18, 20, 20, 25, 25, 30, 30

Your interquartile range is 3 - 25

25 - 3 = 21

8 0

3 years ago

Density is mass divided by volume. Find the density of an aluminum block with a mass of 35 grams and a volume of 14 cm3?

castortr0y [4]

The answer is 2.5 g/cm 3 :)

6 0

4 years ago