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3 years ago

Explain how circle C with center at (−9, 8) and radius 7 is similar to circle D with center at (−1, −1) and radius 5.

Mathematics

1 answer:

Korolek [52]3 years ago

6 0

Answer:

We are told that circle C has center (-4, 6) and a radius of 2.

We are told that circle D has center (6, -2) and a radius of 4.

If we move circle C's center ten units to the right and eight units down, the new center would be at (-4 + 10), (6 - 8) = (6, -2). So step 1 in the informal proof checks out - the centers are the same (which is the definition of concentric) and the shifts are right.

Let's look at our circles. Circle C has a radius of 2 and is inside circle D, whose radius is 4. Between Circle C and Circle D, the radii have a 1:2 ratio, as seen below:

If we dilate circle C by a factor of 2, it means we are expanding it and doubling it. Our circle has that 1:2 ratio, and doubling both sides gives us 2:4. The second step checks out.

Translated objects (or those that you shift) can be congruent, and dilated objects are used with similarity (where you stretch and squeeze). The third step checks out.

Thus, the argument is correct and the last choice is best.

Cute copy and paste:

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Kevin bought 4 pizzas to share with friends. If each person has 2/3 of a pizza how many friends can have pizza?

rjkz [21]

Divide 4 pizzas by 2/3:

When you divide using a fraction, reverse the second fraction and multiply:

4 / 2/3 = 4 x 3/2 = 12/2 = 6 people.

7 0

3 years ago

A model for the movement of a stock supposes that if the present price of the stock is s, then, after one period, it will be eit

aleksklad [387]

Answer:

$P(X\geq 470)=0.9991$

Step-by-step explanation:

Present price of stock is s, if x is the times that the stock increases (by u) among the 1000 periods we can calculate the price after 1000 periods as follows:

$s*u^n*d^{(1000-x)}$ note that if the price increase n periods, so it should decreases in the other 1000-x periods.

We need that the price be 30% higher or 1.3 times the initial price. That requirement is expressed as follows:

$s*u^n*d^{(1000-x)}>1.3s$

applying properties of exponents and reorganizing.

$d^{1000}(\frac{u}{d})^x>1.3$

solving for x

$1000 log(d)+ x log (\frac{u}{d})>log(1.3)$

$x >\frac{log(1.3)-1000 log(d)}{log (\frac{u}{d})}$

Using the values given d= 0.99 and u=1.012

$x > 469.208967$

So, we need at least 470 increase periods

Now we can calculate the probability of have 470 increase periods or more, taking into account that the distribution of x is binomial, the period can be and increase period with 0.52 of probability and we have 1000 total trials.

The formula for the binomial cumulative probability function is

$F(x;p,n) = \Sigma_0^x \left(\begin{array}{c}n&i\end{array}\right)p^i(1-p)^{n-i}$

With this formula we can calculate the probability of obtain x successful trials or less ( $P(X\leq x)$ ). There, p is the probability of success (probability of increase period in this case), x is the maximum number of successful trials, n the number of total trials. We should calculate F with x=470, p=0.52 and n= 1000. The result is $P(X\leq x)$ , but we need $P(X\geq x)$ so we can calculate that probability as