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Dahasolnce [82]
3 years ago
15

You are dispatched to a​ 1-year-old child with respiratory distress. En​ route, you review how to assess and treat infants with

respiratory problems. Which of the following would NOT be a standard part of assessment for​ breathing?
A.
Work of breathing
B.
Mental status
C.
Chest expansion
D.
Sounds of breathing
Biology
1 answer:
Anastasy [175]3 years ago
8 0

Answer:

B. Mental status

Explanation:

Respiratory diseases are medical conditions that affect the lungs and breathing capacity, but do not alter the child's mental state, so in assessing the child's breathing, it will not be necessary to evaluate breathing work, chest expansion and sounds of the child's breath. It is not necessary to evaluate the mental states to know if the child has breathing problems.

Some breathing problems are genetic while others are caused by lifestyle or environmental factors. Common breathing problems include asthma, bronchitis, emphysema, tuberculosis and sinusitis.

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Two main ingredients in plant fertilizer are phosphorus and nitrogen. these elements are found in which classes of biomolecues?
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<u>Nucleic Acid, and Protein</u> are the class of biomolecule which are the two main ingredients in plant fertilizer.

Biomolecules- A chemical substance that is present in living things is called a biomolecule. These consist mostly of substances with the chemical elements carbon, hydrogen, oxygen, nitrogen, sulfur, and phosphorus. The building blocks of life, biomolecules serve crucial roles in all living things.

Fertilizer- Any product or material given to soil to encourage plant development is referred to as fertilizer. There are many different types of fertilizers, and the majority of them include potassium, phosphorus, and nitrogen (N) (K). In actuality, the package of fertilizers bought in supermarkets lists the N-P-K ratio.

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2 years ago
What happens to water (H2O) and carbon dioxide (CO2) during photosynthesis?
Andrew [12]

Answer:

Carbon dioxide turns into glucose and water turns into oxygen

5 0
3 years ago
Read 2 more answers
Please answer the picture
Oksana_A [137]
2, 4, 1, 3, 5

the tiles go into the sequence in that order.

first one is tile 2, second is tile 4, etc.
4 0
3 years ago
Homologous structures are defined as anatomical structures originating from the
ValentinkaMS [17]

Explanation:

An example of homologous structures in vertebrates is

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7 0
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In tomato plants, the allele for purple stems is dominant to allele for green stems and the allele for red fruit is dominant to
Kamila [148]

So let's call the allele that is dominant for purple stems is P, and the recessive allele yielding green stems is p. And let's say that the allele that is dominant for red fruit is R and the recessive allele for yellow fruit is r.


The two plants are each heterozygour for both traits, meaning that the cross is going to be:


PpRr x PpRr


So let's setup a Punnet Square:


\left[\begin{array}{ccccc}&PR&Pr&pR&pr&PR&PPRR&PPRr&PpRR\4&PpRr&Pr&PPRr&PPrr&PpRr&Pprr\\pR&PpRR&PpRr&ppRR&ppRr&pr&PpRr&Pprr&ppRr&pprr\end{array}\right]


The genotypic ratios (alleles) and phenotypic ratios are as follows:


PPRR (Purple stem, Red fruit): \frac{1}{16}


PpRR (Purple stem, Red fruit): \frac{2}{16} or \frac{1}{8}


ppRR (Green Stem, Red fruit): \frac{1}{16}


PPRr (Purple stem, Red fruit): \frac{2}{16} or \frac{1}{8}


PPrr (Purple stem, Yellow fruit): \frac{1}{16}


PpRr (Purple stem, Red fruit): \frac{4}{16} or \frac{1}{4}


pprr (Green stem, Yellow fruit): \frac{1}{16}


ppRr (Green stem, Red fruit): \frac{2}{16} or \frac{1}{8}


Pprr (Purple stem, Yellow fruit): \frac{2}{16} or \frac{1}{8}


So then we take the like phenotypic ratios and add them together:


Purple stem, Red fruit: \frac{1}{16} +\frac{2}{16} +\frac{2}{16} +\frac{4}{16}=\frac{9}{16}


Green stem, Red fruit: \frac{1}{16}+ \frac{2}{16} =\frac{3}{16}


Purple stem, Yellow fruit: \frac{1}{16}+ \frac{2}{16} =\frac{3}{16}


Green stem, Yellow fruit: \frac{1}{16}

6 0
3 years ago
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