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azamat
3 years ago
10

The helmet-like region of the sperm that is similar to a large lysosome and assists penetration of the egg is called the _______

_.
Biology
1 answer:
V125BC [204]3 years ago
8 0
Hello there!

The helmet-like region of the sperm that is similar to a large lysosomr and assists penetration of the egg is called the Acrosome.

The correct term would be "Acrosome".

Hope this helps!

~Lexa
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Option B.

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Explanation:

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What is the sample variance of the following data, which describe the height growth of 10 foxtal pine trees grown in a common ga
maksim [4K]

Answer:

Although the data in the question is not clearly stated, the correct date is stated in the image attached to this answer and it is also stated below

What is the sample variance of the following data, which describe the height growth of 10 foxtail pine trees grown in a common garden ( give your  answer to one decimal place)?

Tree            Height(cm)

1                  25.6987

2                 19.6587

3                 11.4780

4                 9.11130

5                 16.8747

6                 14.7771

7                 33.0001

8                 24.0000

9                 19.4441

10                14.1010

answer 51.9

Explanation:

what is a sample variance?

A sample variance (denoted as S^{2}) is used to calculate how far a set of numbers are from the mean value of the set. Mathematically, it is defined as the average of the squared deviations from its mean value.

Let the individual value of the trees be represented as x₁ to x₁₀

Let the total number of Foxtail pine trees be N

To calculate the standard deviation, we will first calculate the mean or average of the height of the trees

1. Mean (μ) = \frac{x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}+x_{7}+x_{8}+x_{9}+x_{10}  }{N}

                  = \frac{25.6987+19.6587+11.4780+9.11130+16.8747+14.7771+33.0001+24.0000+19.4441+14.1010}{10}  = \frac{188.1437}{10} = 18.8144

2. we need to calculate the sum of the squares of the individual deviations.

∑[x-μ]²= (25.6987-18.8144)^{2}+(19.6587-18.8144)^{2}+(11.4780-18.8144)^{2}+(9.11130-18.8144)^{2}+(16.8747-18.8144)^{2}+(14.7771-18.8144)^{2}+(33.0001-18.8144)^{2}+(24.0000-18.8144)^{2}+(19.4441-18.8144)^{2}+(14.1010-18.8144)^{2}=466.8786

3. Now the variance is ; (∑[x-μ]²) ÷ (N-1) and since N=10, then N-1 = 10-1 = 9

= \frac{466.8786}{9} = 51.8754.

we were asked the give our answer to one decimal place

∴ Variance(S^{2})= 51.9

   

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