Answer:
A. 14.59
Explanation:
Firstly, rounding a number means making a number simpler but keeping its value close. So rounding 14.587020 means making it look less complex but making sure the rounded value is within range of the original value.
Next, we have to know that to round a number;
* Digits below 5 (<5) rounds down the next digit to their left i.e. we leave the digits the same and turn the rounding digit to 0.
* Digits above 5 (>5) rounds up the next digit to the left i.e. we add 1 to the next digit
Next, to round to 4 significant digits means we want to see only the first four non-zero digits in our value. Note that, all non-zero digits i.e. 1-9 are significant
Hence, 14.587020 will be:
= 14.587020
= 14.58700
= 14.59000
We have four (4) digits which are non-zero already, hence we stop here and remove all zeros to the right
Hence, 14.587020 to 4 significant digits is 14.59 i.e. 14.587020 ~ 14.59
The plants can be classified based on their vascular and reproductive structures. The plant given in the image belongs to the Bryophyta.
<h3>What is Bryophyta?</h3>
The non-vascular plants that are a part of the plant classification are found in moist places and are small in size belongs to the Bryophyta.
The plant of this classification have a leafy multicelled plant body, that lacks true leaves, roots, and stems. They are green plants and also contains chloroplast for food synthesis.
The vascular tissue like xylem and phloem are completely absent and hence the transportation does not occur through specialized systems.
Therefore, option A. Bryophyta is the plant growing on the rock.
Learn more about Bryophyta here:
brainly.com/question/1435423
Answer:
B. AUU CGC AUC UUG AAC
Explanation:
<em>According to the rule of base pairing in DNA, the pyrimidine bases always pair with purine bases. Specifically, adenine (A) always pair with thymine (T) while Guanine (G) always pair with cytosine (C). In RNA, thymine is replaced by uracil (U).</em>
Hence, TAA GCG TAG AAC TTG will give
AUU CGC AUC UUG AAC.
The correct option is B.
Given: The population density of frogs in a portion of a pond is 12 frogs per square meter. The size of the entire pond is 370 square meters.
To find: The total population size of the frogs in the pond.
Method: Using unitary method of proportions
Solution:
Let p be the density of the frogs present per square meter.
Let A denote the total area of the pond.
∴ p = 12 frogs/m² and A = 370 m²
The total number of frogs in the pond is given by,
Therefore, there are 4440 frogs in the pond.
