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Nostrana [21]
2 years ago
15

Insect Weights Consider a dataset giving the adult weight of species of insects. Most species of insects weigh less than 5 grams

, but there are a few species that weigh a great deal, including the largest insect known: the rare and endangered Giant Weta from New Zealand, which can weigh as much as 71 grams. Is the shape of the distribution symmetric, skewed to the right, or skewed to the left
Mathematics
1 answer:
Ann [662]2 years ago
8 0

Answer:

Step-by-step explanation:

71 grams would definitely be an outlier on the high side, whereas "most" species would weigh much less.  Thus, the graph of this distribution of weights would be skewed towards the lower side, that is, to the left.

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A rectangular container can hold 45 candles, while a cylindrical container holds 41 candles. Each rectangular
ki77a [65]

Answer:

143

157

The rectangular containers because it will cost the company $67 more to use cylindrical containers

Step-by-step explanation:

5 0
2 years ago
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Two standardized​ tests, a and​ b, use very different scales of scores. the formula upper a equals 40 times upper b plus 50a=40×
Romashka [77]

Answer:

(a) The mean score of test A is 1210.

(b) The mean score of test A is 1210.

(c) The standard deviation of  test A is 80.

(d) The value of Q₃ for test A is 1170.

(e) The value of median for test A is 1090.

(f) The value of IQR for test A is 290.

Step-by-step explanation:

The relation between two standardized tests <em>A</em> and <em>B</em> is:

50A=40B+50

The equation above approximates the relationship between scores on the two tests.

The summary statistics for test B are as follows:

Lowest Score = 21

Mean score = 29

Standard deviation = 2

Q₃ = 28

Median = 26

IQR = 6

(a)

Compute the lowest score on test A as follows:

A=40B+50\\=(40\times21)+50\\=890

Thus, the lowest score on test A is 890.

(b)

Compute the mean score of test A as follows:

<h2>A=40B+50\\=(40\times29)+50\\=1210</h2>

Thus, the mean score of test A is 1210.

(c)

Compute the mean score standard deviation of test A as follows:

<h2>A=40B+50\\=(40\times2)\\=80</h2>

Thus, the standard deviation of  test A is 80.

(d)

Compute the value of Q₃ for test A as follows:

A=40B+50\\=(40\times28)+50\\=1170

Thus, the value of Q₃ for test A is 1170.

(e)

Compute the value of median for test A as follows:

A=40B+50\\=(40\times26)+50\\=1090

Thus, the value of median for test A is 1090.

(f)

Compute the value of IQR for test A as follows:

A=40B+50\\=(40\times6)+50\\=290

Thus, the value of IQR for test A is 290.

6 0
2 years ago
Wanda is shopping for a pet carrier. One small carrier can hold 240 ounces. Her cat weighs 12 pounds. Can the carrier hold her c
daser333 [38]
Yes, 12 x 16 gives us the amount of ounces the cat weighs which is 192 ounces. Therefore, her cat CAN fit. 192 oz < 240 oz
4 0
2 years ago
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Using the information given, select the statement that can deduce the line segments to be parallel. If there are none, then sele
Alina [70]

we know that

<u>1) If the line segment AB is parallel to the line segment DC</u>

then

m∠1=m∠D -------> by corresponding angles

<u>2) If the line segment AD is parallel to the line segment BC</u>

then

m∠3=m∠B -------> by corresponding angles

5 0
2 years ago
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The annual salaries of all employees at a financial company are normally distributed with a mean of $34,000 and a standard devia
Alla [95]

Answer:

z=5

Step-by-step explanation:

We have been given that the The annual salaries of all employees at a financial company are normally distributed with a mean of $34,000 and a standard deviation of $4,000.

To solve our given problem we will use z-score formula.

z=\frac{x-\mu}{\sigma} where,

z=\text{z-score},

x=\text{Sample score},

\mu=\text{Mean},

\sigma=\text{Standard deviation}

Upon substituting our given values in z-score formula we will get,

z=\frac{54,000-34,000}{4,000}

z=\frac{20,000}{4000}

z=5

Therefore, the z-score of a company employee who makes an annual salary of $54,000 is 5.

4 0
2 years ago
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