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Phantasy [73]
3 years ago
9

Which type of symptoms negatively affect the intellectual functioning of an individual with schizophrenia?

Biology
1 answer:
White raven [17]3 years ago
5 0
Schizophrenia is a mental disorder defined as chronic and severely debilitating as the patients usually are "lost in touch with reality."

Schizophrenia has both positive symptoms (added) and negative symptoms (lack of). Positive symptoms include hallucinations (<em>adding </em>unreal imagery) and uncontrollable thoughts. Negative symptoms include apathy <em />(<em>lack of </em>concern or enthusiasm), alogia (<em>lack of </em>speech), and blunted affect (<em>lack of </em>facial expression). Both positive and negative symptoms affect the intellectual functioning of an individual with schizophrenia. 
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A force of 2.20 N is exerted on a 7,20 g rifle bullet. What is the bullet's acceleration? (Enter your answer in m/s^ 2 )
sweet [91]

Answer:

The answer is 306 m/s^2

Explanation:

F = m*a

We can rearrange this formula to get a = F / m.

The SI unit for mass is kilograms (kg).

The SI unit for acceleration is m/s^2.

There are 1000 grams in a kilogram.

7.20 g = 0.00720 kg

Force (2.20 N) / 0.00720 kg = 305.56 m/s^2

Accounting for Sig Figs, the answer is 306 m/s^2.

Gee, that's one fast bullet.

7 0
3 years ago
In a population of 150 toads, poisonous (P) is dominant over non-poisonous (p). According to the Hardy-Weinberg equation, the ex
aalyn [17]

Answer: Χ²=8.18

Explanation: In this population, the frequency of homozygote poisonous toads is:

p^{2} =\frac{100}{150}

p^{2} = 0.667

p = \sqrt{0.667}

p=0.82

For the homozygote non-poisonous toads:

q^{2} = \frac{5}{150}

q^{2} =0.034

q = \sqrt{0.034}

q = 0.18

The frequency for heterozygote poisonous toad, we can use the Hardy-Weinberg equation: p^{2} + 2pq + q^{2} = 1, in which, heterozygote frequency is given by 2pq

2pq = 2*0.82*0.18 = 0.3

Now, to compute the chi-square test, follow the instructions:

1) Find the observed values: in this case, they are the found frequency:

0.82       0.3         0.18

2) Find the expected values: As the question mentioned, the expected proportion is:

16            8             1

3) Subtract the observed value from the expected value:

0.82 - 16 = - 0.184          0.3 - 8 = -7.7             0.18 - 1 = - 0.818

4) Square each value from above:

(-0.184)² = 0.034            (-7.7)² = 59.3                 (-0.818)² = 0.77

5) Divide each value by expected value:

\frac{0.034}{16} = 0.0021                   \frac{59.3}{8} = 7.41                      \frac{0.77}{1} = 0.77

6) Add all the values and we will have the chi-square test:

Χ² = 0.0021 + 7.41 + 0.77 = 8.18

The chi-square test is Χ² = 8.18

8 0
3 years ago
Urgent help
34kurt

Archaebacteria = 10,000/20 = 500 species  

Eubacteria = 10,000/1 = 10,000 species  

Do the rest the same way, since you've got the graph right in front of you.

7 0
3 years ago
What is the smallest organized unit of life?
Rainbow [258]
A cell. it is the smallest unit of life that can reproduce indepentantly
4 0
4 years ago
If a population has an allele A1 frequency of 0.04 and an allele A2 frequency of 0.96, what is the expected frequency of the A2A
torisob [31]

Answer:

A2A2 = 0.9216

Explanation:

Due to technical problems, you will find the complete answer and explanation in the attached files.    

Download pdf
6 0
3 years ago
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