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MrMuchimi
3 years ago
8

When researching online, as long as the gathered information is the result of a valid Internet search engine, it is not necessar

y to analyze it for bias.
Biology
1 answer:
Hitman42 [59]3 years ago
8 0
It is always necessary to analyze research for bias
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Which of the following statements about fungi is not true? a. All fungi are useful for plant growth b. Athletes foot is caused b
Vera_Pavlovna [14]

Answer:

c

Explanation:

a is true , b athletes foot is caused by a type of fungi d fungi true since they are like magots.

5 0
3 years ago
Segun la informacion que contenga el nucleo¿Como se pueden clasificar las celulas?​
Digiron [165]

Answer:

En biología celular, el núcleo (pl. Núcleos; del latín nucleus o nuculeus, que significa núcleo o semilla) es un orgánulo unido a la membrana que se encuentra en las células eucariotas.

Explanation:

8 0
3 years ago
Removing which of the following parts would cause the plant to starve? Leaves Branches Stems Roots
daser333 [38]
I believe it would be the leaves because that helps collect water! :)
8 0
3 years ago
Two species of spiny mice compete for the same food source. One species of spiny mice feeds on insects during the day, while the
Korvikt [17]

Answer:

Temporal partitioning or niche differentiation

Explanation:

Temporal partitioning refers to the variability in the competitive abilities of different species competing for the same resources based on varying conditions of the environment.

The environmental factors could be light, temperature, moisture, etc. Each species specializes in feeding within varied condition of one or more of the factors.

<em>In this case, both species are in competition but while the first species feeds during the day, the second species feeds during the night. Here, light and temperature probably plays a huge factor.</em>

Temporal partitioning is also known as niche differentiation, segregation, separation or niche partitioning.

3 0
3 years ago
Each child born to a particular set of parents has probability 0.25 of having blood type o. if these parents have 5 children, wh
Taya2010 [7]
Use binomial distribution with p=0.25, n=5, x=2
P(X=x)=C(n,x)p^x (1-p)^(n-x)

P(X=2)=C(5,2) 0.25^2 0.75^(5-2)
=10*0.0625*0.421875
=0.26367

So the probability of 2 of 5 children having type-O blood from these parents is 0.26367.

7 0
3 years ago
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