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lara31 [8.8K]
3 years ago
15

9u²v and which term are like terms?

Mathematics
1 answer:
MaRussiya [10]3 years ago
4 0
There are no like terms in 9u^2v because it is all one term and none of the variables are like one another.
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The graph below represents the following system of inequalities. y>x-2 y>2x+2 which points (x,y) satisfies the given syste
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Answer: (-4,-6) is the point that ALMOST satisfies both inequalities. IF they were equalities, this would be the solution.

The question is a bit confusing as it asks for "which points (x,y) satisfies both" It's ungrammatical, and many points (infinite within the shaded region) are solutions that SATISFY the system of inequalities!

Step-by-step explanation: Substitute the x and y-values and see if the inequalities are true.

y>x-2 -6> -4-2 -6= -6

That point (-4,-6) is on the dashed line, so not exactly a true solution; this is a question about inequalities. So y values have to be greater than-6 or x-values less than -4 for a true inequality.

y>2x+2

-6>(2)(-4) +2

-6> -8 +2

-6> -6 Again, equal, so for this y-values have to be greater than-6 and/or x-values less than -4 in order to have a true inequality.

If you have the graph to look at, you can select any points in the shaded region that satisfies both of the inequalities.

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In a fitness centre, the monthly fee is charged at $55 per hour. For a member of the fitness centre, when the charge in a monthl
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Answer:

The clothes you wear for working out at a gym will impact your performance. You must choose performance-oriented and high-quality workout clothes that are breathable and don't restrict your movement. You must wear clothes that are stretchable and offer the right fit.

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Step-by-step explanation:

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A horticulturist working for a large plant nursery is conducting experiments on the growth rate of a new shrub. Based on previou
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Answer:

t=\frac{1.8-2}{\frac{0.5}{\sqrt{48}}}=-2.771    

The degrees of freedom are given by:

df=n-1=48-1=47  

And the p value would be:

p_v =P(t_{(47)}    

Since the p value is lower than the significance level we have enough evidence to conclude that the true mean for this case for the growth rate is less than 2cm per week

Step-by-step explanation:

Information given

\bar X=1.8 represent the sample mean for the growth

s=0.5 represent the sample standard deviation    

n=48 sample size    

\mu_o =2 represent the value that we want to compare

\alpha=0.01 represent the significance level

t would represent the statistic    

p_v represent the p value

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is less than 2cm per week, the system of hypothesis are :    

Null hypothesis:\mu \geq 2    

Alternative hypothesis:\mu < 2    

Since we don't know the population deviation the statistic is given by:

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

Replacing the info given we got:

t=\frac{1.8-2}{\frac{0.5}{\sqrt{48}}}=-2.771    

The degrees of freedom are given by:

df=n-1=48-1=47  

And the p value would be:

p_v =P(t_{(47)}    

Since the p value is lower than the significance level we have enough evidence to conclude that the true mean for this case for the growth rate is less than 2cm per week

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