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2 years ago

108 identical books have a mass of 30 kg. Find

Mathematics

1 answer:

julia-pushkina [17]2 years ago

8 0

Answer:

The mass of 150 books = 41.7kg

The number of books that have mass 20kg = 72 books

Step-by-step explanation:

(I) Number of identical books = 108

Mass of books = 30 kg

Mass of 150 books = ?

Let the mass of 150 books = x kg

To find the mass of 150 books we will use ratio and proportion:

108 books: 30kg = 150 books: x kg

You can also write it like this because ratio can be written in division:

108/30 = 150/x

By cross multiplication:

108x = 150*30

108 x= 4500

Divide both sides by 108

108 x/108 = 4500/108

x= 41.7

Thus the mass of 150 books = 41.7kg

( ii ) The number of books that have a mass of 20 kg = ?

Let assume that the number of books of mass 20 kg = y books

We will again use ratio and proportion method.

108 books: 30kg = y: 20kg

We can write it as:

108/30 = y/20

By cross multiplication:

108 *20 = 30y

2160= 30y

Divide both the terms by 30

2160/30 = 30y/30

72 = y

Therefore the number of books that have mass 20kg = 72 books ....

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Answer: The conditional statements are not in the correct form to make a conclusion using the law of syllogism. “If p, then q and if p, then r” cannot be used to draw a conclusion using the law of syllogism. The law of syllogism could be used if the hypothesis in the second statement was "if two pairs of congruent angles are formed."

Step-by-step explanation:

"If p, then q and if p, then r" cannot be used to draw a conclusion using the law of syllogism.

Neither of the conclusions of the conditional statements are the hypothesis of the other.

"If two pairs of congruent angles are formed" could be the hypothesis of the second statement.

** Both can be used to answer the question :)

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1 year ago

Read 2 more answers

A shoe manufacturer compared material A and material B for the soles of shoes. Twelve volunteers each got two shoes. The left wa

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Answer:

a) Are dependent since we are mesuring at the same individuals but on different times and with a different method

b) If we see the qq plot we don't have any significant deviation for the values and we don't have any heavy tail so we can conclude that we can approximate the differences with the normal distribution.

c) $p_v =P(t_{(12)}>0.969) =0.353$

So the p value is higher than the significance level given, so then we can conclude that we FAIL to reject the null hypothesis. So we can conclude that the mean differences is NOT significantly different from 0 .

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.

The Q-Q plot, or quantile-quantile plot, "is a graphical tool to help us assess if a set of data plausibly came from some theoretical distribution such as a Normal or exponential".

Let put some notation

x=value for A , y = value for B

A: 379, 378, 328, 372, 325, 304, 356, 309, 354, 318, 355, 392

B: 372, 376, 328, 368, 283, 252, 369, 321, 379, 303, 328, 411

(a) Are the two samples paired or independent? Explain your answer.

Are dependent since we are mesuring at the same individuals but on different times and with a different method

(b) Make a normal QQ plot of the differences within each pair. Is it reasonable to assume a normal population of differences?

The first step is calculate the difference $d_i=A_i-B_i$ and we obtain this:

d: 7,2,0,4,42,52,-13,-12,-25,15,27,-19

In order to do the qqplot we can use the following R code:

d<-c(7,2,0,4,42,52,-13,-12,-25,15,27,-19)

qqnorm(d)

And the graph obtained is attached.

If we see the qq plot we don't have any significant deviation for the values and we don't have any heavy tail so we can conclude that we can approximate the differences with the normal distribution.

(c) Choose a test appropriate for the hypotheses above and justify your choice based on your answers to parts (a) and (b). Perform the test by computing a p-value, make a test decision, and state your conclusion in the context of the problem

The system of hypothesis for this case are:

Null hypothesis: $\mu_A- \mu_B = 0$

Alternative hypothesis: $\mu_A -\mu_B \neq 0$

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{80}{12}=6.67$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =23.849$

The 4 step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{6.67 -0}{\frac{23.849}{\sqrt{12}}}=0.969$

The next step is calculate the degrees of freedom given by:

$df=n-1=12-1=11$

Now we can calculate the p value, since we have a two tailed test the p value is given by:

$p_v =P(t_{(12)}>0.969) =0.353$

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2 years ago

Two horses are ready to return to their barn after a long workout session at the track. the horses are at coordinates H(1,10) an

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The given information are the coordinates of points H, Z and B. So, the first step to do is to plot these points on a Cartesian plane as shown in the attached picture. We can deduce visually that horse Z is closer to the barns than horse H. But, to further justify the answer, we have to provide the magnitude of the distance of horses H and Z to barn B. In this approach, we use the distance formula:

d = √(x₂ - x₁)² + (y₂ - y₁)²

Use the coordinates of the two points to know their linear distances. These are represented by the red and green lines for each of the horses.

Distance between H and B = √(⁻3 - 1)² + (⁻9-10)²
Distance between H and B = √(⁻3 - 1)² + (⁻9-10)²
Distance between H and B = 19.4165

Since the scale is 1 unit = 100 meters, the actual distance between horse H and barn B is 19.416*100 = 1,941.65 meters

Distance between Z and B = √(⁻3 - 10)² + (⁻9-1)²
Distance between Z and B = √(⁻3 - 10)² + (⁻9-1)²
Distance between Z and B = 16.4012

Since the scale is 1 unit = 100 meters, the actual distance between horse Z and barn B is 16.4012*100 = 1,640.12 meters

Comparing the distances: 1,941.65 meters > 1,640.12 meters. Therefore, it justifies that horse Z is nearer to the barn.