Two organisms of different species are not likely to compete for the same

Which situation would most likely be a driving force for changes in a species?

vovikov84 [41]

A situation which is most likely to be a driving force for changes in a species is increased mating events due to environmental stability.

<h3>What is a species?</h3>

A species can be defined as a biological classification of related organisms that have similar characteristics and they are capable of mating (breeding) with one another in a cycle or different seasons.

This ultimately implies that, increased mating events between living organisms due to environmental stability is a situation which is most likely to be a driving force for changes in a species.

Read more on species here: brainly.com/question/19878144

7 0

2 years ago

For a laboratory assignment if the equipment is working the density function of the observed outcome

vivado [14]

Based the density function of the observed outcome, the probability $P(x\leq \frac{1}{3} )$ is equal to 0.1388.

<h3>What is a density function?</h3>

A density function can be defined as a type of function which is used to represent the density of a continuous random variable that lies within a specific range.

Given the following density function:

$f(x)=\left \{ {{2(1-x),\;0 < x < 1} \atop {0,\; otherwise}} \right.$

Next, we would calculate $P(x\leq \frac{1}{3} )$ ;

$P(x\leq \frac{1}{3} )=\int\limits^\frac{1}{3} _0 {f(x)} \, dx \\\\P(x\leq \frac{1}{3} )=2 \int\limits^\frac{1}{3} _0 {(1-x)} \, dx \\\\P(x\leq \frac{1}{3} )=2|x-\frac{x^2}{2} |\limits^\frac{1}{3} _0\\\\P(x\leq \frac{1}{3} )=2[\frac{1}{3} -\frac{1}{2} (\frac{1}{3})^2]\\\\P(x\leq \frac{1}{3} )=2[\frac{1}{3} -\frac{1}{18}]\\\\P(x\leq \frac{1}{3} )=0.1388$

Read more on probability here: brainly.com/question/25870256

4 0

2 years ago

education and career focused opportunity that introduces students to the link between classes in careers is called a(n)

Naya [18.7K]

School to work program is an education, an career focused opportunity

7 0

3 years ago

A dilation produces a smaller figure. Which is a possible scale factor? –2 one-third 1 4.

Sedbober [7]

Answer:

we know that

if a dilation produces a smaller figure

then

0< scale factor <1

so

case A) -2------> the scale factor must be a positive number

case B) 1/3----> 0< 1/3 <1---> is a possible scale factor

case C) 1------> this scale factor does not not produce a smaller figure, it produces the same figure

case D) 4----> this scale factor produces a larger figure

the answer is

1/3

Explanation:

please mark branlyist if correct

7 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