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3 years ago

What figure is a dilation of Figure A by a factor of 1/2 ?

Mathematics

2 answers:

lora16 [44]3 years ago

7 0

A dilation of figure A by a factor of 1/2 is the last figure with sides 1in, 2 in, 0.5 in and 2.5 inch. Since dilation means becoming larger, smaller or wider from one figure, this is also same with similar figures where the ratios of corresponding sides should be the same. In this problem, the ratio is 0.5 or 1/2. Therefore figure A is similar to the last figure.

Ahat [919]3 years ago

5 0

Answer: The answer is (D) the last figure.

Step-by-step explanation: We are given a figure A with sides of measurement a = 2 in, b = 5 in, c = 1 in and d = 4 in.. The figure A is dilated by a factor of $\dfrac{1}{2}$ . We need to select the correct figure after dilation out of the given four options.

Since the figure A is dilated by a scale factor of $\dfrac{1}{2}$ , so the new figure will have the following measurement of sides:

$a'=\dfrac{1}{2}\times 2=1~\textup{in.},\\\\\\b'=\dfrac{1}{2}\times 5=2.5~\textup{in.},\\\\\\c'=\dfrac{1}{2}\times 1=0.5,~\textup{in.}\\\\\\d'=\dfrac{1}{2}\times 4=2~\textup{in.}.$

Therefore, the correct figure is the last one (D). The correct figure is attached herewith.

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galben [10]

Answer:

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3 years ago

Paige launched a ball using a catapult she built. The height of the ball (in meters above the ground) ttt seconds after launch i

agasfer [191]

Answer:

1) Factor form : $h(t)=-5t(t-8)$

2) 8 second after launch.

Step-by-step explanation:

The height of the ball (in meters above the ground) t seconds after launch is modeled by

$h(t)=-5t^2+40t$

To find the time when ball hit the ground, we need to find the factor form of the given function.

$h(t)=-5t(t-8)$

When ball hi the ground, then height of the ball from the ground is 0.

$h(t)=0$

$-5t(t-8)=0$

Using zero product property, we get

$-5t=0\Rightarrow t=0$

$t-8=0\Rightarrow t=8$

Ball hit the ground at t=0 and t=8. It means ball hit the ground in starting and 8 second after launch.

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3 years ago

7. In right triangle BCD, BD12, mLC 90, andmLDBC47. Find DC to the nearest tenth47

tresset_1 [31]

This is using sine law

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3 years ago

Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean

Mamont248 [21]

Answer:

a) The percentage of snails that take more than 60 hours to finish is 4.75%.

b) The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c) The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d) 0% probability that a randomly-chosen snail will take more than 76 hours to finish

e) To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f) The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 50, \sigma = 6$

a. The percentage of snails that take more than 60 hours to finish is

This is 1 subtracted by the pvalue of Z when X = 60.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

1 - 0.9525 = 0.0475

The percentage of snails that take more than 60 hours to finish is 4.75%.

b. The relative frequency of snails that take less than 60 hours to finish is

This is the pvalue of Z when X = 60.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c. The proportion of snails that take between 60 and 67 hours to finish is

This is the pvalue of Z when X = 67 subtracted by the pvalue of Z when X = 60.

X = 67

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{67 - 50}{6}$

$Z = 2.83$

$Z = 2.83$ has a pvalue 0.9977

X = 60

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

0.9977 - 0.9525 = 0.0452

The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d. The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places)

This is 1 subtracted by the pvalue of Z when X = 76.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{76 - 50}{6}$

$Z = 4.33$

$Z = 4.33$ has a pvalue of 1

1 - 1 = 0

0% probability that a randomly-chosen snail will take more than 76 hours to finish

e. To be among the 10% fastest snails, a snail must finish in at most hours.

At most the 10th percentile, which is the value of X when Z has a pvalue of 0.1. So it is X when Z = -1.28.

$Z = \frac{X - \mu}{\sigma}$

$-1.28 = \frac{X - 50}{6}$

$X - 50 = -1.28*6$

$X = 42.32$

To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f. The most typical 80% of snails take between and hours to finish.

From the 50 - 80/2 = 10th percentile to the 50 + 80/2 = 90th percentile.

10th percentile

value of X when Z has a pvalue of 0.1. So X when Z = -1.28.

$Z = \frac{X - \mu}{\sigma}$

$-1.28 = \frac{X - 50}{6}$

$X - 50 = -1.28*6$

$X = 42.32$

90th percentile.

value of X when Z has a pvalue of 0.9. So X when Z = 1.28

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 50}{6}$

$X - 50 = 1.28*6$

$X = 57.68$

The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

5 0

3 years ago

La barbería El Caleño, tiene en promedio 120 clientes a la semana a

Luba_88 [7]

Queremos maximizar el precio de tal forma que los ingresos no disminuyan.

Ese maximo precio es: $14,040.6

Sabemos que actualmente el precio es:

p = $6,000

El número de clientes es:

C = 120

Actualmente los ingresos son el producto de esos dos números, es decir:

ingresos = $6,000*120 = $720,000

Ahora sabemos que por cada incremento de $700 en el precio, el número de clientes decrece en 10.

Entonces podemos escribir el número de clientes como una ecuación lineal.

C(p) = a*p + b

tal que tenemos dos puntos en esa linea:

($6,000, 120)

($6,700, 110)

La pendiente es:

$a = \frac{110 - 120}{\$6,700 - \$6,000} = \frac{-10}{\$ 700}$

Entonces tenemos:

C(p) = (-10/$700)*p + b

Sabemos que:

C($6,000) = 120 = (-10/$700)*$6,000 + b

120 = -85.71 + b

120 + 85.71 = b =

Entonces la ecuación lineal es:

C(p) = (-10/$700)*p + 205.71

Los ingresos serán dados por:

ingresos = C(p)*p = (-10/$700)*p^2 + 205.71*p

Y queremos maximizar p de tal forma que esto sea igual a lo que obtuvimos antes:

(-10/$700)*p^2 + 205.71*p = $720,000

Entonces debemos resolver la ecuación cuadratica:

(-10/$700)*p^2 + 205.71*p - $720,000 = 0.

Las soluciones son dadas por la formula de Bhaskara.

$p = \frac{-205.71 \pm \sqrt{(205.71)^2 - 4*(-10/\$ 700)*\$ 720,000} }{2*(-10/\$ 700)} \\\\p = \frac{-205.71 \pm 195.45}{(-20/\$ 700)}$

La solución de maximo valor es:

p = (-205.71 - 195.45)/(-20/$700) = $14,040.6

Sí quieres aprender más, puedes leer.

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2 years ago