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2 years ago

The question is 56 : 28 as 122 : ?

Mathematics

2 answers:

Gnom [1K]2 years ago

8 0

56/28 = 122/x
56x = 122(28)
56x = 3416
56x/56 = 3416/56
x = 61

56:28 as 122:61

RUDIKE [14]2 years ago

6 0

Answer:

56 : 28 as 122 : 61

Step-by-step explanation:

Proportion says that two ratios or fractions are equal.

As per the statement:

56 : 28 :: 122 : ?

Let x be the fourth proportion.

then;

56 : 28 :: 122 : x

by definition of proportion:

$\frac{56}{28}=\frac{122}{x}$

by cross multiply we have;

$56x = 3,416$

Divide both sides by 56 we have;

x = 61

Therefore, 56 : 28 as 122 : 61

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High-power experimental engines are being developed by the Stevens Motor Company for use in its new sports coupe. The engineers

mamaluj [8]

Answer:

The 95% confidence interval the average maximum power is (596.0 to 644.0)

Step-by-step explanation:

Average maximum of the sample = x = 620 HP

Standard Deviation = s = 45 HP

Sample size = n = 16

We have to calculate the 95% confidence interval. The value of Population standard deviation is unknown, and value of sample standard deviation is known. Therefore, we will use one sample t-test to build the confidence interval.

Degrees of freedom = df = n - 1 = 15

Critical t-value associated with 95% confidence interval and 15 degrees of freedom, as seen from t-table = $t_{\frac{\alpha}{2}}$ = 2.131

The formula to calculate the confidence interval is:

$(x-t_{\frac{\alpha}{2} } \times \frac{s}{\sqrt{n} }, x+t_{\frac{\alpha}{2} } \times \frac{s}{\sqrt{n} })$

We have all the required values. Substituting them in the above expression, we get:

$(620-2.131 \times \frac{45}{\sqrt{16} }, 620+2.131 \times \frac{45}{\sqrt{16} })\\\\ =(596.0 , 644.0)$

Thus, the 95% confidence interval the average maximum power is (596.0 to 644.0)

7 0

2 years ago

How many pairs of whole numbers have a sum of 99

alexandr402 [8]

With the sum of 99, we will get 50 pairs whole numbers. Why?

Let’s start with

0+ 99

1 + 98

2 + 97

3 + 96

4 + 95

5 + 94

6 + 93

7 + 92

8 + 91

9 + 90

10 + 89

………

……..

43 + 49

44 + 50

Therefore, if you’re going to count all pairs of whole number, you will get 50 pairs of whole number with the sum of 99.

Hope this helps!

4 0

2 years ago

Read 2 more answers

(Pic above) In the figure above , O is the center of square. What is the area of the square?

azamat

The answer is 64

- the side length of the square is 8 units
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1 year ago

An experiment on memory was performed, in which 16 subjects were randomly assigned to one of two groups, called "Sentences" or "

FromTheMoon [43]

Answer:

There is no significant difference between the averages.

Step-by-step explanation:

Let's call

$\large X_{sentences}$ the mean of the “sentences” group

$\large S_{sentences}$ the standard deviation of the “sentences” group

$\large X_{intentional}$ the mean of the “intentional” group

$\large S_{intentional}$ the standard deviation of the “intentional” group

Then, we can calculate by using the computer

$\large X_{sentences}=28.75$

$\large S_{sentences}=3.53553$

$\large X_{intentional}=31.625$

$\large S_{intentional}=1.40788$

$\large X_{sentences}-X_{intentional}=28.75-31.625=-2.875$

The standard error of the difference (of the means) for a sample of size 8 is calculated with the formula

$\large \sqrt{(S_{sentences})^2/8+(S_{intentional})^2/8}$

So, the standard error of the difference is

$\large \sqrt{(3.53553)^2/8+(1.40788)^2/8}=1.34546$

In order to see if there is a significant difference in the averages of the two groups, we compute the interval of confidence of 95% for the difference of the means corresponding to a level of significance of 0.05 (5%).

If this interval contains the zero, we can say there is no significant difference.

Since the sample size is small, we had better use the Student's t-distribution with 7 degrees of freedom (sample size-1), which is an approximation to the normal distribution N(0;1) for small samples.

We get the $\large t_{0.05}$ which is a value of t such that the area under the Student's t distribution outside the interval $\large [-t_{0.05}, +t_{0.05}]$ is less than 0.05.