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2 years ago

If f(x) = √2x+3 6x-5 a. 1 b. -2 then f() = C. -1 d. - 13

Mathematics

2 answers:

Gekata [30.6K]2 years ago

8 0

Answer:

put 1/2as a value of x in the equation and then solve .then the answer will be -1

blsea [12.9K]2 years ago

7 0

Answer:

$C.\ f\left( \frac{1}{2} \right) =-1$

Step-by-step explanation:

$f\left( x\right) =\frac{\sqrt{2x+3} }{6x-5}$

In order to calculate f(1/2) ,all we have to do

is replacing x by 1/2 in the expression of f.

Then

$f\left( \frac{1}{2} \right) =\frac{\sqrt{2\left( \frac{1}{2} \right) +3} }{6\left( \frac{1}{2} \right) -5 }$

$=\frac{\sqrt{1+3} }{3-5}$

$=\frac{\sqrt{4} }{-2}$

$=\frac{2}{-2}$

$=-1$

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Find the area of the shaded regions below. Give your answer as a completely

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Answer:

You just need to find the length of LM and NM and then find the area of semi-circles using the formula πr²÷2

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3 years ago

Assume that the Poisson distribution applies to the number of births at a particular hospital during a randomly selected day. As

kakasveta [241]

Answer:

0.9999985 = 99.99985% probability that in a day, there will be at least 1 birth.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Assume that the mean number of births per day at this hospital is 13.4224.

This means that $\mu = 13.4224$

Find the probability that in a day, there will be at least 1 birth.

This is:

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-13.4224}*13.4224^{0}}{(0)!} = 0.0000015$

Then

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3 years ago

1. Consider an athlete running a 40-m dash. The position of the athlete is given by , where d is the position in meters and t is

sasho [114]

There is some information missing in the question, since we need to know what the position function is. The whole problem should look like this:

Consider an athlete running a 40-m dash. The position of the athlete is given by $d(t)=\frac{t^{2}}{6}+4t$ where d is the position in meters and t is the time elapsed, measured in seconds.

Compute the average velocity of the runner over the intervals:

(a) [1.95, 2.05]

(b) [1.995, 2.005]

(d) [2, 2.00001]

Answer

(a) 6.00041667m/s

(b) 6.00000417 m/s

(d) 6.00001 m/s

The instantaneous velocity of the athlete at t=2s is 6m/s

Step by step Explanation:

In order to find the average velocity on the given intervals, we will need to use the averate velocity formula:

$V_{average}=\frac{d(t_{2})-d(t_{1})}{t_{2}-t_{1}}$

so let's take the first interval:

(a) [1.95, 2.05]

$V_{average}=\frac{d(2.05)-d(1.95)}{2.05-1.95}$

we get that:

$d(1.95)=\frac{(1.95)^{3}}{6}+4(1.95)=9.0358125$

$d(2.05)=\frac{(2.05)^{3}}{6}+4(2.05)=9.635854167$

so:

$V_{average}=\frac{9.6358854167-9.0358125}{2.05-1.95}=6.00041667m/s$

(b) [1.995, 2.005]

$V_{average}=\frac{d(2.005)-d(1.995)}{2.005-1.995}$

we get that:

$d(1.995)=\frac{(1.995)^{3}}{6}+4(1.995)=9.30335831$

$d(2.005)=\frac{(2.005)^{3}}{6}+4(2.005)=9.363335835$

so:

$V_{average}=\frac{9.363335835-9.30335831}{2.005-1.995}=6.00000417m/s$

$V_{average}=\frac{d(2.0005)-d(1.9995)}{2.0005-1.9995}$

we get that:

$d(1.9995)=\frac{(1.9995)^{3}}{6}+4(1.9995)=9.33033358$

$d(2.0005)=\frac{(2.0005)^{3}}{6}+4(2.0005)=9.33633358$

so:

$V_{average}=\frac{9.33633358-9.33033358}{2.0005-1.9995}=6.00000004m/s$

(d) [2, 2.00001]

$V_{average}=\frac{d(2.00001)-d(2)}{2.00001-2}$

we get that:

$d(2)=\frac{(2)^{3}}{6}+4(2)=9.33333333$

$d(2.00001)=\frac{(2.00001)^{3}}{6}+4(2.00001)=9.33339333$

so:

$V_{average}=\frac{9.33339333-9.33333333}{2.00001-2}=6.00001m/s$

Since the closer the interval is to 2 the more it approaches to 6m/s, then the instantaneous velocity of the athlete at t=2s is 6m/s

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3 years ago

Read 2 more answers