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Anettt [7]
3 years ago
5

add or subtract the fractions for the addition problems write another addition problem that has the same sum and use two or more

fractions 9/12 - 2/12
Mathematics
1 answer:
lesya692 [45]3 years ago
6 0
You can add these fractions to get 9/12 - 2/12 ( 7/12 )

3/12 + 4/12
1/12 + 6/12
2/12 + 5/12

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Esearchers believed that an increase in lean body mass is associated with an increase in maximal oxygen uptake. A scatterplot of
Dovator [93]

Answer:

The correct option is (D).

Step-by-step explanation:

The <em>p</em>-value is well defined as per the probability, [under the null hypothesis (<em>H₀</em>)], of attaining a result equivalent to or more extreme than what was truly observed.

A small <em>p</em>-value (typically ≤ 0.05) specifies solid proof against the null hypothesis (<em>H₀</em>), so you discard <em>H₀</em>.

A large <em>p</em>-value (> 0.05) specifies fragile proof against the <em>H₀</em>, so you fail to discard <em>H₀</em>.

The hypothesis is defined as follows:

<em>H₀</em>: The slope of the regression line is 0, i.e. <em>β </em>= 0.

<em>Hₐ</em>: The slope of the regression line is greater than 0, i.e. <em>β </em>> 0.

A scatter-plot of the measurements taken from 18 randomly selected college athletes displayed a strong positive linear relationship between the two variables; lean body mass and maximal oxygen uptake.

The significance level of the test is, <em>α</em> = 0.05.

The <em>p</em>-value of the test is, <em>p</em>-value = 0.04

As the <em>p</em>-value = 0.04 < <em>α</em> = 0.05, we reject the null hypothesis.

So, it can be concluded that there is a statistical evidence of a relationship between lean body mass and maximal oxygen uptake for college athletes.

Since the scatter-plot shows a strong positive linear relationship, it implies that as the an increase in lean body mass causes an increase in maximal oxygen uptake for college athletes.

Thus, the correct option is (D).

7 0
3 years ago
Suppose that it is known that the number of items produced in a factory during a week is a random variable with mean 50. If the
iVinArrow [24]

Answer:

The probability would be 0.9586

Step-by-step explanation:

Given,

Mean, \mu = 50,

Variance, \sigma^2 = 24

Standard deviation,

\sigma = \sqrt{24}=2\sqrt{6},

If X represents the number of items produced in a factory during a week.

Hence, the probability that this weeks production will be between 40 and 60

=P(40 < X < 60)

=P(\frac{40-\mu}{\sigma} < \frac{X-\mu}{\sigma} < \frac{60-\mu}{\sigma})

=P(\frac{40-50}{2\sqrt{6}} < z < \frac{60-50}{2\sqrt{6}})

=P(-\frac{10}{2\sqrt{6}} < z < \frac{10}{2\sqrt{6}})

=P(-2.041 < z < 2.041)

=P(z< 2.041) - P(z< -2.041)

Using z score table

= 0.9793 - 0.0207

= 0.9586

4 0
3 years ago
-0.5 x 1/3 estimate each product or quotient and then compute
Gnoma [55]
 -0.5 * 0.3 =  - 0.15 .
______________________________________
7 0
3 years ago
Solve for y ; -2y - 2x =4
Deffense [45]
I think this is what you want to do

Your welcome

4 0
3 years ago
Read 2 more answers
Work out the value of six squared plus three cubed
IrinaK [193]

Answer:

63

Step-by-step explanation:

Based off of the question,we have a word problem that can be interpreted as the following. 6²+3³= ?

6² is the same as 6×6 which is equal to 36, In other words 6×6=36 or 6²=36

3³ is the same as 3×3×3 which is equal to 27, In other words 3×3×3=27 or

3³=27

Now with the product of each value in the word problem we can set up a statement as the following: 36+27= ?

Which when added together gives us the sum of 63, in other terms 36+27=63

concluding in our answer for the initial word problem. (6²+3³=63)

7 0
3 years ago
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