There are 5,280 feet in a mile. What part of a mile, in decimal form, will you drive until you reach the exit.

Tomas is a commercial real estate agent who earns a yearly salary of $82,500 and a 2.5% commission on his total sales. If Tomas

Georgia [21]

Tomas earning is $176,250

7 0

1 year ago

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iven: C is a point on the perpendicular bisector, l, of AB. Prove: AC = BC Use the drop-down menus to complete the proof. By the

ZanzabumX [31]

Answer:

See the attached figure for better explanation :

Step-by-step explanation :

1. By the unique line postulate, you can draw only one line segment : BC

Since only one line can be drawn between two distinct points.

2. Using the definition of reflection, reflect BC over l.

To find line segment which reflects BC over l, we will use the definition of reflection.

3. By the definition of reflection, C is the image of itself and A is the image of B.

Definition of reflection says the figure about a line is transformed to form the mirror image. Now, CD is perpendicular bisector of AB so A and B are equidistant from D forming the mirror image of each other.

4. Since reflections preserve length, AC = BC

In Reflection the figure is transformed to form a mirror image, Hence the length will be preserved in case of reflection.

8 0

2 years ago

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In a (blank),one ratio compares a part to a whole

Verdich [7]

Answer:

In a part-to-whole ratio, one ratio compares a part to a whole.

3 0

2 years ago

Victor fue al mercado para comprar manzanas, naranjas y platanos; las naranjas costaron el doble de lo 1ue pago por las manzanas

Thepotemich [5.8K]

Answer:

El precio de las manzanas = 27 pesos

El precio de las naranjas = 54 pesos

El precio de las bananas = 19 pesos

Step-by-step explanation:

Los parámetros dados son;

El monto total gastado = 100 pesos

Sea el precio de las naranjas = x

Sea el precio de las manzanas = y

Sea el precio de los plátanos = z

La cantidad pagada por las naranjas = 2 · y = x

La cantidad pagada por los plátanos = y - 8 = z

Por lo tanto, tenemos;

La cantidad total gastada = La cantidad pagada por las naranjas + La cantidad pagada por las bananas + La cantidad pagada por las manzanas

∴ El monto total gastado = 100 pesos = 2 · y + y - 8 + y

100 = 4 · años - 8

4 · y = 100 + 8 = 108

y = 108/4 = 27

y = 27

De

z = y - 8 tenemos;

z = 27 - 8 = 19

De 2 · y = x, tenemos;

2 × 27 = x

x = 54

Por lo tanto;

El precio de las naranjas = 54 pesos

El precio de las manzanas = 27 pesos

El precio de los plátanos = 19 pesos.

5 0

1 year ago

Let Y denote a geometric random variable with probability of success p. a Show that for a positive integer a, P(Y > a) = qa .

lakkis [162]

Answer:

a) For this case we can find the cumulative distribution function first:

$F(k) = P(Y \leq k) = \sum_{k'=1}^k P(Y =k')= \sum_{k'=1}^k p(1-p)^{k'-1}= 1-(1-p)^k$

So then by the complement rule we have this:

$P(Y>a) = 1-F(a)= 1- [1-(1-p)^a]= 1-1 +(1-p)^a = (1-p)^a = q^a$

b) $P(Y>a)= q^a$

$P(Y>b) = q^b$

So then we have this using independence:

$P(Y> a+b) = q^{a+b}$

We want to find the following probability:

$P(Y> a+b |Y>a)$

Using the definition of conditional probability we got:

$P(Y> a+b |Y>a)= \frac{P(Y> a+b \cap Y>a)}{P(Y>a)} = \frac{P(Y>a+b)}{P(Y>a)} = \frac{q^{a+b}}{q^a} = q^b = P(Y>b)$

And we see that if a = 2 and b=5 we have:

$P(Y> 2+5 | Y>2) = P(Y>5)$

c) For this case we use independent identical and with the same distribution experiments.

And the result for part b makes sense since we are interest in find the probability that the random variable of interest would be higher than an specified value given another condition with a value lower or equal.

Step-by-step explanation:

Previous concepts

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

$P(X=x)=(1-p)^{x-1} p$

If we define the random of variable Y we know that:

$Y\sim Geo (1-p)$

Part a

For this case we can find the cumulative distribution function first:

$F(k) = P(Y \leq k) = \sum_{k'=1}^k P(Y =k')= \sum_{k'=1}^k p(1-p)^{k'-1}= 1-(1-p)^k$

So then by the complement rule we have this:

$P(Y>a) = 1-F(a)= 1- [1-(1-p)^a]= 1-1 +(1-p)^a = (1-p)^a = q^a$

Part b

For this case we can use the result from part a to conclude that:

$P(Y>a)= q^a$

$P(Y>b) = q^b$

So then we have this assuming independence:

$P(Y> a+b) = q^{a+b}$

We want to find the following probability:

$P(Y> a+b |Y>a)$

Using the definition of conditional probability we got:

$P(Y> a+b |Y>a)= \frac{P(Y> a+b \cap Y>a)}{P(Y>a)} = \frac{P(Y>a+b)}{P(Y>a)} = \frac{q^{a+b}}{q^a} = q^b = P(Y>b)$

And we see that if a = 2 and b=5 we have:

$P(Y> 2+5 | Y>2) = P(Y>5)$

Part c

For this case we use independent identical and with the same distribution experiments.

And the result for part b makes sense since we are interest in find the probability that the random variable of interest would be higher than an specified value given another condition with a value lower or equal.

8 0

2 years ago