Plzzz help. Giving everything.

Adriana y Luisa fueron a la feria y decidieron subirse a la rueda de la fortuna. Si el diámetro de la rueda es de 8 metros, ¿Qué

sashaice [31]

Answer:

Translation?

Step-by-step explanation:

6 0

3 years ago

Solve the system y = -x + 7 and y = 0.5(x - 3)2

amid [387]

Answer:

(5, 2 ) and (- 1, 8 )

Step-by-step explanation:

Given the equations

y = - x + 7 → (1)

y = 0.5(x - 3)²

= 0.5(x² - 6x + 9)

y = 0.5x² - 3x + 4.5 → (2)

Substitute y = - x + 7 into (2)

- x + 7 = 0.5x² - 3x + 4.5 ( subtract - x + 7 from both sides )

0 = 0.5x² - 2x - 2.5 ( multiply through by 2 ) , then

x² - 4x - 5 = 0 ← in standard form

(x - 5)(x + 1) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 5 = 0 ⇒ x = 5

x + 1 = 0 ⇒ x = - 1

Substitute these values into (1) for corresponding values of y

x = 5 : y = - 5 + 7 = 2 ⇒ (5, 2 )

x = - 1 : y = -(- 1) + 7 = 1 + 7 = 8 ⇒ (- 1, 8 )

6 0

3 years ago

How to convert degrees minutes seconds to decimal degrees by hand?

Thepotemich [5.8K]

Multiply secondsn by 60 add to minutes multiply by 60 and to degrees

5 0

3 years ago

Suppose that prices of a certain model of new homes are normally distributed with a mean of $150,000. use the 68-95-99.7 rule to

Novosadov [1.4K]

Answer:

We want to find the percentage of values between 147700 and 152300

$P(147700$

And one way to solve this is use a formula called z score in order to find the number of deviations from the mean for the limits given:

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

And replacing we got:

$z=\frac{147700-150000}{2300}=-1$

$z=\frac{152300-150000}{2300}=1$

So then we are within 1 deviation from the mean so then we can conclude that the percentage of values between $147,700 and $152,300 is 68%

Step-by-step explanation:

We define the random variable representing the prices of a certain model as X and the distirbution for this random variable is given by:

$X \sim N(\mu = 150000, \sigma =2300$

The empirical rule states that within one deviation from the mean we have 68% of the data, within 2 deviations from the mean we have 95% and within 3 deviations 99.7 % of the data.

We want to find the percentage of values between 147700 and 152300

$P(147700$

And one way to solve this is use a formula called z score in order to find the number of deviations from the mean for the limits given:

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

And replacing we got:

$z=\frac{147700-150000}{2300}=-1$

$z=\frac{152300-150000}{2300}=1$

So then we are within 1 deviation from the mean so then we can conclude that the percentage of values between $147,700 and $152,300 is 68%

7 0

3 years ago

On the following number line, the approximate locations of five real numbers are graphed.

NemiM [27]

Answer:

C

A

E

D

B

I got this right on my assignment

5 0

4 years ago