All categories

2 years ago

Point out the characteristics of the musical style.

SAT

1 answer:

katrin2010 [14]2 years ago

5 0

Answer:

Francisco bue Camino

Francisco Santiago

nicanor abelardo

Antonia Molina

hilarion Rubio

col Antonio Buenaventura

rodalfa cornejo

Felipe Padilla de Leon SR

rosendo Santos jr

Alfredo Buenaventura

Ayan cayabyab

You might be interested in

A polling agency is investigating the voter support for a ballot measure in an upcoming city election. The agency will select a

Rudiy27

Based on the information given, it should be noted that the probability that the proportion of voters that will support the ballot measure will be greater than 0.50 is 0.0895.

<h3>How to calculate the probability.</h3>

From the information given, it was stated that a polling agency is investigating the voter support for a ballot measure in an upcoming city election and that the the population proportion of voters who would support the ballot measure in region a is 0. 47.

Based on the complete information, the probability that the proportion of voters in the sample of region A that will support the ballot measure will be greater than 0.50 will be given as P(Z > 1.3441).

It should be noted that when this value is looked in the table, it will give a value of 0.0895.

Learn more about probability on:

brainly.com/question/25870256

4 0

3 years ago

Factor the expression into an equivalent form 12y^2-75.

elena-14-01-66 [18.8K]

Hello oddworld7836!

$\huge \boxed{\mathbb{QUESTION} \downarrow}$

Factor the expression into an equivalent form 12y² - 75.

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$12 y ^ { 2 } - 75$

By observing the expression, we can see that, 3 is the only common factor in both the terms of the expression. So, take the common factor 3 out.

$12 y ^ { 2 } - 75 \\ = 3\left(4y^{2}-25\right)$

Now, look at (4y² - 25). They don't have any common factors but they appear in the form of the algebraic identity ⇨ a² - b² = (a + b) (a - b). Here,

a² = 4, a = 2 (√a² = ✓4 = 2)
b² = 25, b = 5 (√b² = ✓25 = 5)

So, the (4y² + 25) becomes...

$(4 {y}^{2} - 25) \\ = \left(2y-5\right)\left(2y+5\right)$

Now, bring the 3 (common factor) & rewrite the complete expression.

$12 y ^ { 2 } - 75 \\ = \boxed{ \boxed{ \bf \: 3\left(2y-5\right)\left(2y+5\right) }}$

We can't further simplify it. Also, remember that the simplified form of an expression is equivalent to the expression. So, 3 (2y - 5) (2y + 5) is equivalent to 12y² - 75.

__________________

Hope it'll help you!

ℓu¢αzz ッ

8 0

2 years ago

In addition to replacing officers lost to desertions, defections, and retirements, what was among the significant rebuilding cha

trasher [3.6K]

People often face one or more problems daily. Police Department faced following Hurricane Katrina is had the challenge of Refurbishing flooded police headquarters.

<h3>The challenges that face police officers during Hurricane Katrina?</h3>

There were a lot of social implications of natural disasters, such as the one of Hurricane Katrina. This was known to be of huge concern to police, because it brought about the shifting duties that centers around crime control and emergency relief to the breakdown or vadalization of some key elements of the police infrastructure.

It also brought about the the destruction of police communication and transportation.

See options below

a. Restoring healthy water supplies

b. Repaving streets and roads

c. Improving city utility services

d. Refurbishing flooded police headquarters

Learn more about hurricane Katrina from

brainly.com/question/332478

5 0

2 years ago

The air bag inflates and deflates so quickly so? A. You will not smother B. You can escape or drive to avoid a collision C. A an

Archy [21]

The best answer is B

4 0

3 years ago

Read 2 more answers

Friction: a 50-kg box is resting on a horizontal floor. A force of 250 n directed at an angle of 30. 0° below the horizontal is

sergejj [24]

The net forces on the box parallel and perpendicular to the surface, respectively, are

∑ F[para] = (250 N) cos(-30.0°) - F[friction] = (50 kg) a

and

∑ F[perp] = F[normal] + (250 N) sin(-30.0°) - F[weight] = 0

where a is the acceleration of the box. (We ultimately don't care what this acceleration is, though.)

To decide whether the friction here is static or kinetic, consider that when the box is at rest, the net force perpendicular to the floor is

∑ F[perp] = F[normal] - F[weight] = 0

so that, while at rest,

F[normal] = (50 kg) g = 490 N

Then with µ[s] = 0.40, the maximum magnitude of static friction would be

F[s. friction] = 0.40 (490 N) = 196 N

so that the box will begin to slide if it's pushed by a force larger than this.

The horizontal component of our pushing force is

(250 N) cos(-30.0°) ≈ 217 N

so the box will move in our case, and we will have kinetic friction with µ[k] = 0.30.

Solve the ∑ F[perp] = 0 equation for F[normal] :

F[normal] + (250 N) sin(-30.0°) - F[weight] = 0

F[normal] - 125 N - 490 N = 0

F[normal] = 615 N

Then the kinetic friction felt by the box has magnitude

F[k. friction] = 0.30 (615 N) = 184.5 N ≈ 185 N

8 0

2 years ago