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alina1380 [7]
2 years ago
15

Based on the graph shown to the right, whose phones cost varies directly with the amount of time spent talking?

Mathematics
2 answers:
Lena [83]2 years ago
8 0

Answer:

B

Step-by-step explanation:

Given 2 quantities that vary directly, then the graph must pass through the origin.

Nikiya's graph is the only one to do this ⇒ B



sineoko [7]2 years ago
6 0

Answer:

B. Nikiya's only

Step-by-step explanation:

Mark’s cost does not vary directly with time spent talking because the cost of 0 hours is not 0 dollars

For Nikiya’s plan, the value of the constant of variation is 15

which represents Nikiya’s cost per hour

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JulsSmile [24]
For this case what you should see is that for the interval [9, 11] the behavior of the function is almost linear.
 Therefore, we can find the average rate of change as follows:
 m = (y2-y1) / (x2-x1)
 m = (11-6) / (11-9)
 m = (5) / (2)
 m = 5/2
 Answer:
 the average rate of speed over the interval [9, 11] is: 
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2 years ago
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The following two sets of parametric functions both represent the same ellipse. Explain the difference between the graphs.
statuscvo [17]
The answer
ellipse main equatin is as follow:

X²/ a²   +  Y²/ b²  =1, where a≠0 and b≠0

for the first equation: <span>x = 3 cos t and y = 8 sin t
</span>we can write <span>x² = 3² cos² t and y² = 8² sin² t
and then  </span>x² /3²= cos² t and y²/8² =  sin² t
therefore,  x² /3²+ y²/8²  =  cos² t + sin² t = 1
equivalent to x² /3²+ y²/8²  = 1

for the second equation, <span>x = 3 cos 4t and y = 8 sin 4t we found
</span>x² /3²+ y²/8²  = cos² 4t + sin² 4t=1

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2 years ago
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Crop researchers plant 15 plots with a new variety of corn. The yields in bushels per acre are: 138.0 139.1 113.0 132.5 140.7 10
son4ous [18]
There is a relationship between confidence interval and standard deviation:
\theta=\overline{x} \pm \frac{z\sigma}{\sqrt{n}}
Where \overline{x} is the mean, \sigma is standard deviation, and n is number of data points.
Every confidence interval has associated z value. This can be found online.
We need to find the standard deviation first: 
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When we do all the calculations we find that:
\overline{x}=123.8\\ \sigma=11.84
Now we can find confidence intervals:
($90\%,z=1.645): \theta=123.8 \pm \frac{1.645\cdot 11.84}{\sqrt{15}}=123.8 \pm5.0\\($95\%,z=1.960): \theta=123.8 \pm \frac{1.960\cdot 11.84}{\sqrt{15}}=123.8 \pm 5.99\\ ($99\%,z=2.576): \theta=123.8 \pm \frac{2.576\cdot 11.84}{\sqrt{15}}=123.8 \pm 7.87\\
We can see that as confidence interval increases so does the error margin. Z values accociated with each confidence intreval also get bigger as confidence interval increases.
Here is the link to the spreadsheet with standard deviation calculation:
https://docs.google.com/spreadsheets/d/1pnsJIrM_lmQKAGRJvduiHzjg9mYvLgpsCqCoGYvR5Us/edit?usp=sharing
6 0
2 years ago
Identify the key characteristics of the parent fifth-root function f (x) = ^5sqrtx. Include the following: domain, range, interv
Andre45 [30]

Given functin is :

f\left(x\right)=\sqrt[5]{x}

We know that the domain of the expression is all real numbers except where the expression is undefined. In given function, there is no real number that makes the expression undefined. Hence domain is all real numbers.

Domain: (-∞,∞)

Range is the set of y-values obtained by plugging values from domain so the range will also same.

Range: (-∞,∞)

If we increase value of x then y-value will also increase so that means it is an INCREASING function. You can also verify that from graph.

It crosses x and y-axes both at the origin

Hence x-intercept=0 and y-intercept=0

Graph is not symmetric about y-axis hence it can't be EVEN

Graph is not symmetric about origin so it is ODD.

There is  no breaking point in the graph so that means it is a Continuous function.

There is no hoirzontal or vertical or slant line which seems to be appearing to touch the graph at infinity so there is NO asymptote.

END behaviour means how y-changes when x approaches infinity.

From graph we can see that when x-approaches -∞ then y also approaches ∞.

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boyakko [2]

Answer:

c. \frac{1}{5}

Step-by-step explanation:

Given,

Number of bacterias who alive for at least 30 days = 5,

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Died bacterias = 10,

Total bacterias = 5 + 10 + 10 = 25,

Ways of choosing a bacteria = ^{25}C_1 = \frac{25!}{1! 24!} = 25,

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Hence, the probability it will still be alive after one week = \frac{5}{25} = \frac{1}{5}

OPTION C is correct.  

3 0
2 years ago
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