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Fofino [41]
3 years ago
13

What is the simplified form of the following expression? 5√8-√18-2√2

Mathematics
1 answer:
Romashka [77]3 years ago
6 0

Answer:

Option (B) is correct.

To prove

As given the epression in the question is given by

= 5\sqrt{8} - \sqrt{18} - 2 \sqrt{2}

Now simplify the above

Terms are written as

\sqrt{8} = \sqrt{2\times 2\times 2} = 2 \sqrt{2}

\sqrt{18} = \sqrt{3\times 3\times 2} = 3 \sqrt{2}

Put in the above expression.

= 5\times 2 \sqrt{2} -3 \sqrt{2} - 2 \sqrt{2}

= 10 \sqrt{2} -5 \sqrt{2}

=5 \sqrt{2}

Therefore\ the\ expression\ 5\sqrt{8} - \sqrt{18} - 2 \sqrt{2}\ is\ equivalent\ to\ 5 \sqrt{2} .





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siniylev [52]
48 + (2*13) / 110
PE(MD)(AS)

2 * 13 = 26
48 + 26 / 110

26 / 110 = 0.236..
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2 years ago
A container built for transatlantic shipping is constructed in the shape of a right rectangular prism. Its dimensions are 7.5 ft
Yuri [45]

Considering the definition of right rectangular prism and its volume,  the total weight of the contents is 38.81 pounds.

<h3>Right rectangular prism</h3>

A right rectangular prism (or cuboid) is a polyhedron whose surface is formed by two equal and parallel rectangles called bases and by four lateral faces that are also parallel rectangles and equal two to two.

<h3>Volume of right rectangular prism</h3>

To calculate the volume of the rectangular prism, it is necessary to find the product of its dimensions, or of the three edges that converge at a certain vertex.

That is, to calculate the volume of a rectangular prism, multiply its 3 dimensions: length×width×height.

<h3>Volume of the container </h3>

In this case, you know that:

  • the dimensions of the container built are 7.5 ft by 11.5 ft by 3 ft.
  • the container is entirely full and, on average, its contents weigh 0.15 pounds per cubic foot.

So, the volume of the container is calculated as:

7.5 ft× 11.5 ft× 3 ft= <u><em>258.75 ft³</em></u>

Then, the total weight of the contents is calculated as:

258.75 ft³×  0.15 pounds per cubic foot= <u><em>38.8125 pounds≅ 38.81 pounds</em></u>

Finally, the total weight of the contents is 38.81 pounds.

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3 years ago
The owner of a produce stand is selling baskets of peaches, priced by the kilogram. She rounded each measurements to the nearest
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Answer:

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Step-by-step explanation:

Given

a = 6\frac{1}{2} --- heaviest

b = 5 --- second heaviest

Required

The difference (d)

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d = 6\frac{1}{2} - 5

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3 years ago
To evaluate the effect of a treatment, a sample is obtained from a population, with a mean of u = 30, and the treatment is admin
Murrr4er [49]

Answer:

a) t=\frac{31.3-30}{\frac{3}{\sqrt{16}}}=1.733  

p_v =2*P(t_{15}>1.733)=0.104  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the we don't have a significant effect for the new treatment at 5% of significance.  

b) t=\frac{31.3-30}{\frac{3}{\sqrt{36}}}=2.6  

p_v =2*P(t_{15}>2.6)=0.0201  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the we have a significant effect for the new treatment at 5% of significance.  

c) When we increase the sample size we increse the probability of rejection of the null hypothesis since the z score tend to increase when the sample size increase.

Step-by-step explanation:

Data given and notation  

Part a: If the sample consists of n=16 individuals, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with alpha = 0.05.

\bar X=31.3 represent the sample mean  

s=3 represent the sample standard deviation  

n=16 sample size  

\mu_o =30 represent the value that we want to test  

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is different from 30, the system of hypothesis are :  

Null hypothesis:\mu = 30  

Alternative hypothesis:\mu \neq 30  

Since we don't know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{31.3-30}{\frac{3}{\sqrt{16}}}=1.733  

P-value  

First w eneed to find the degrees of freedom given by:

df=n-1=16-1 =15

Since is a two-sided test the p value would given by:  

p_v =2*P(t_{15}>1.733)=0.104  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the we don't have a significant effect for the new treatment at 5% of significance.  

Part b: If the sample consists of n=36 individuals, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with alpha = 0.05.

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{31.3-30}{\frac{3}{\sqrt{36}}}=2.6  

P-value  

First w eneed to find the degrees of freedom given by:

df=n-1=16-1 =15

Since is a two-sided test the p value would given by:  

p_v =2*P(t_{15}>2.6)=0.0201  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the we have a significant effect for the new treatment at 5% of significance.  

Part c; Comparing your answer for parts a and b, how does the size of the sample influence the outcome of a hypothesis test

When we increase the sample size we increse the probability of rejection of the null hypothesis since the z score tend to increase when the sample size increase.

5 0
3 years ago
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