Simplify the following expression: 5(y + 12) - 34 Help me please! (15 points)

Square root of 19.4 rounded to the nearest hundredth

photoshop1234 [79]

Answer:

there is none.

Step-by-step explanation:

19.4, the lowest is tenths, you need another digit down for hundredths.

hope I helped!

5 0

3 years ago

se tiene que embaldosar el patio interior de un edificio con baldosas cuadradas de 30 cm de lado. El patio es rectangular y sus

shepuryov [24]

Answer:

40,000 baldosas

Step-by-step explanation:

Lo primero que debemos hacer aquí es calcular el área del patio rectangular.

El mejor enfoque para esto es convertir primero sus medidas a centímetros

Matemáticamente, 100 cm = 1 m, entonces 10 m se convierten en 1000 cm y 12 m se convierten en 1200 m.

El área de un rectángulo es L * B y, por lo tanto, tenemos 1200 * 1000 = 1,200,000 cm ^ 2

Ahora, para saber la cantidad de azulejos que tendrá el patio, necesitaremos dividir el área del patio por el área de los azulejos

Matemáticamente, eso sería 1,200,000 / 30 = 40,000 fichas

6 0

3 years ago

A baker makes 84 loaves of bread, she burns 4. What fraction of the loaves does the burn? Simplify your answer.

svetoff [14.1K]

Answer:

1/21

Step-by-step explanation:

The baker makes 84 loaves of bread,

Out of the 84 bread, 4 of them got burn.

Fraction that the loaves does the burn will be expressed in 4/84 = 1/21.

3 0

3 years ago

Given f(x) = 5x - 1 what is f(-2)?

Finger [1]

Answer:

-11

Step-by-step explanation:

Plug in -2 for x

5(-2) - 1

Order of operations (or PEMDAS) tells you to multiply first giving you:

-10 - 1

-10 - 1 = -11

3 0

3 years ago

The heights of women in the USA are normally distributed with a mean of 64 inches and a standard deviation of 3 inches.

Rainbow [258]

Answer:

(a) 0.2061

(b) 0.2514

(c) 0

Step-by-step explanation:

Let X denote the heights of women in the USA.

It is provided that X follows a normal distribution with a mean of 64 inches and a standard deviation of 3 inches.

(a)

Compute the probability that the sample mean is greater than 63 inches as follows:

$P(\bar X>63)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{63-64}{3/\sqrt{6}})\\\\=P(Z>-0.82)\\\\=P(Z$

Thus, the probability that the sample mean is greater than 63 inches is 0.2061.

(b)

Compute the probability that a randomly selected woman is taller than 66 inches as follows:

$P(X>66)=P(\frac{X-\mu}{\sigma}>\frac{66-64}{3})\\\\=P(Z>0.67)\\\\=1-P(Z$

Thus, the probability that a randomly selected woman is taller than 66 inches is 0.2514.

(c)

Compute the probability that the mean height of a random sample of 100 women is greater than 66 inches as follows:

$P(\bar X>66)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{66-64}{3/\sqrt{100}})\\\\=P(Z>6.67)\\\\\ =0$

Thus, the probability that the mean height of a random sample of 100 women is greater than 66 inches is 0.

8 0

3 years ago