A car traveled an average speed of 60mph for two hours. how far did it travel?

What is the simplified form of the following expression 2 sqrt 27 + sqrt 12 -3 sqrt 3-2 sqrt 12

Dmitrij [34]

Answer:

√3

Step-by-step explanation:

2√27 + √12 − 3√3 − 2√12

2√27 − √12 − 3√3

2√(9×3) − √(4×3) − 3√3

2×3√3 − 2√3 − 3√3

6√3 − 2√3 − 3√3

√3

3 0

3 years ago

Suppose that the halflife of a substance is 250 years . If there were ir 100 the substance , (a) give an exponential model for t

IgorLugansk [536]

After 500 years one would have N nuclei of the original substance.

N = 100 * exp ( -ln(4) ) = 25

LOL.

I guess it is exp( - lamda * t)

Lamda, the decay constant is the ln(2) over the half life.

So we get exp( -ln(2)/250 * 500) or exp ( -ln(4) )
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Related Questions (More Answers Below)

5 0

3 years ago

URGENT!!! PLEASE LOOK AT THE PICTURE!!! PLEASE HELP!!!!

Svetllana [295]

Answer:

Step-by-step explanation:

this is a 45-45-90 triangle. The angle opposite the 45° angle is also 45°.

HI = HJ because this is an isosceles triangle.

Use the cos function to find the length of HJ:

cos Ф = adj / hyp or adj = (hyp)(cos Ф).

1

Here we have adj = HI = HJ = 4√5 cos 45° = 4√5 · -----

√2, or 4√5 / √2

4√10

This last result simplifies to ----------

2

or 2√10 (answer)

The length of side HJ and that of HI is 2/√10.

7 0

4 years ago

The average THC content of marijuana sold on the street is 9.3%. Suppose the THC content is normally distributed with standard d

velikii [3]

Answer:

a) $X \sim N(9.3,1)$

b) $P(X>9.2)=P(\frac{X-\mu}{\sigma}>\frac{9.2-\mu}{\sigma})=P(Z>\frac{9.2-9.3}{1})=1-P(Z$

c) The value of height that separates the bottom 75% of data from the top 25% is 9.9745.

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

2) Part a

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

$X \sim N(9.3,1)$

Where $\mu=9.3$ and $\sigma=1$

3) Part b

We are interested on this probability

$P(X>9.2)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>9.2)=P(\frac{X-\mu}{\sigma}>\frac{9.2-\mu}{\sigma})=P(Z>\frac{9.2-9.3}{1})=1-P(Z$

4) Part c

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.25$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.75 of the area on the left and 0.25 of the area on the right it's z=0.6745. On this case P(Z<0.6745)=0.75 and P(z>0.6745)=0.25

If we use condition (b) from previous we have this:

$P(X$