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inessss [21]
3 years ago
11

If Doris paid $12.80 for 6.4 pounds of Swiss cheese, what was the price of 1

Mathematics
1 answer:
m_a_m_a [10]3 years ago
7 0

Answer:

2

Step-by-step explanation:

12.80=6.4x

12.8/6.4=2

x=2

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The concentration C of certain drug in a patient's bloodstream t hours after injection is given by
frozen [14]

Answer:

a) The horizontal asymptote of C(t) is c = 0.

b) When t increases, both the numerator and denominator increases, but given that the grade of the polynomial of the denominator is greater than the grade of the polynomial of the numerator, then the concentration of the drug converges to zero when time diverges to the infinity. There is a monotonous decrease behavior.  

c) The time at which the concentration is highest is approximately 1.291 hours after injection.

Step-by-step explanation:

a) The horizontal asymptote of C(t) is the horizontal line, to which the function converges when t diverges to the infinity. That is:

c = \lim _{t\to +\infty} \frac{t}{3\cdot t^{2}+5} (1)

c = \lim_{t\to +\infty}\left(\frac{t}{3\cdot t^{2}+5} \right)\cdot \left(\frac{t^{2}}{t^{2}} \right)

c = \lim_{t\to +\infty}\frac{\frac{t}{t^{2}} }{\frac{3\cdot t^{2}+5}{t^{2}} }

c = \lim_{t\to +\infty} \frac{\frac{1}{t} }{3+\frac{5}{t^{2}} }

c = \frac{\lim_{t\to +\infty}\frac{1}{t} }{\lim_{t\to +\infty}3+\lim_{t\to +\infty}\frac{5}{t^{2}} }

c = \frac{0}{3+0}

c = 0

The horizontal asymptote of C(t) is c = 0.

b) When t increases, both the numerator and denominator increases, but given that the grade of the polynomial of the denominator is greater than the grade of the polynomial of the numerator, then the concentration of the drug converges to zero when time diverges to the infinity. There is a monotonous decrease behavior.  

c) From Calculus we understand that maximum concentration can be found by means of the First and Second Derivative Tests.

First Derivative Test

The first derivative of the function is:

C'(t) = \frac{(3\cdot t^{2}+5)-t\cdot (6\cdot t)}{(3\cdot t^{2}+5)^{2}}

C'(t) = \frac{1}{3\cdot t^{2}+5}-\frac{6\cdot t^{2}}{(3\cdot t^{2}+5)^{2}}

C'(t) = \frac{1}{3\cdot t^{2}+5}\cdot \left(1-\frac{6\cdot t^{2}}{3\cdot t^{2}+5} \right)

Now we equalize the expression to zero:

\frac{1}{3\cdot t^{2}+5}\cdot \left(1-\frac{6\cdot t^{2}}{3\cdot t^{2}+5} \right) = 0

1-\frac{6\cdot t^{2}}{3\cdot t^{2}+5} = 0

\frac{3\cdot t^{2}+5-6\cdot t^{2}}{3\cdot t^{2}+5} = 0

5-3\cdot t^{2} = 0

t = \sqrt{\frac{5}{3} }\,h

t \approx 1.291\,h

The critical point occurs approximately at 1.291 hours after injection.

Second Derivative Test

The second derivative of the function is:

C''(t) = -\frac{6\cdot t}{(3\cdot t^{2}+5)^{2}}-\frac{(12\cdot t)\cdot (3\cdot t^{2}+5)^{2}-2\cdot (3\cdot t^{2}+5)\cdot (6\cdot t)\cdot (6\cdot t^{2})}{(3\cdot t^{2}+5)^{4}}

C''(t) = -\frac{6\cdot t}{(3\cdot t^{2}+5)^{2}}- \frac{12\cdot t}{(3\cdot t^{2}+5)^{2}}+\frac{72\cdot t^{3}}{(3\cdot t^{2}+5)^{3}}

C''(t) = -\frac{18\cdot t}{(3\cdot t^{2}+5)^{2}}+\frac{72\cdot t^{3}}{(3\cdot t^{2}+5)^{3}}

If we know that t \approx 1.291\,h, then the value of the second derivative is:

C''(1.291\,h) = -0.077

Which means that the critical point is an absolute maximum.

The time at which the concentration is highest is approximately 1.291 hours after injection.

5 0
3 years ago
What is an equation of the line that is parallel to y=4x−10 and passes through (1, 13) ? Enter your equation in the box.
Evgen [1.6K]

y = 4x−10 ; slope m = 4

Parallel has same slope so slope of new line = 4

y = 4x + b

b = y - 4x

Passes through (1, 13)

b = 13 - 4(1)

b = 13 - 4

b = 9

Equation

y = 4x + 9

3 0
3 years ago
Please answer number three I give brainliest thank y’all so much!
mestny [16]

Answer:

Step-by-step explanation:

2 and 6

5 0
3 years ago
Read 2 more answers
Olivia's team played 8 games of basketball. During those 8 games her team's score was: 70, 67, 79, 66, 64, 80, 64 and 75. Find t
Setler [38]

Answer:

70

Step-by-step explanation:

First put it to least to greatest- 64, 64, 66, 67, 70, 75, 80

Then find the median which is the number in the middle

Round it so it gives us are answer of 70

8 0
3 years ago
g If the p-value for your test statistic in a test of hypothesis (for example, a t-test, one-tail) is 0.07 and the significance
Reptile [31]

Complete Question

If the p-value for your test statistic in a test of hypothesis (for example, a t-test, one-tail) is 0.07 and the significance level alpha is set at 0.05, you conclude that:

A  

Both hypotheses H0 and H1 can be accepted

B

The null hypothesis H0 cannot be rejected

C

The null hypothesis H0 can be rejected

D

The research hypothesis H1 can be accepted

Answer:

The correct option is B

Step-by-step explanation:

From the question we are told that

     The p-value  is  p-value  =  0.07

     The level of significance is  \alpha  =  0.05

Generally given that it is a one tail test then the possible value of  

   The null hypothesis is  H_o : \mu =  x

Here \mu is the mean and  x  is the value  

 the possible values of the  alternative hypothesis is  \mu >  x  \ \ or  \ \  \mu < x

Now from the value given we see that the  p-value >  \alpha  hence

The decision rule is  

    Fail to reject  the null hypothesis

       

7 0
3 years ago
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