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Amanda [17]
1 year ago
6

Liam works at a zoo. He was looking at some data showing the masses of their 555 African elephants. The mean mass of the elephan

ts was 3{,}800\,{\text{kg}}3,800kg3, comma, 800, start text, k, g, end text, and the median mass was 3{,}600\,{\text{kg}}3,600kg3, comma, 600, start text, k, g, end text. The smallest elephant, named Lola, weighed 2{,}700\,{\text{kg}}2,700kg2, comma, 700, start text, k, g, end text. Lola then got very sick and lost weight until her mass reached 1{,}800\,{\text{kg}}1,800kg1, comma, 800, start text, k, g, end text. How will Lola's mass decreasing affect the mean and median? Choose 1 answer: Choose 1 answer: (Choice A) A Both the mean and median will decrease. (Choice B) B The median will decrease, and the mean will stay the same. (Choice C) C The mean will decrease, and the median will stay the same. (Choice D) D The mean will decrease, and the median will increase.
Mathematics
3 answers:
My name is Ann [436]1 year ago
6 0

a Step-by-step explanation:

Westkost [7]1 year ago
6 0

Answer:

The answer is C

Step-by-step explanation:

Guest1 year ago
0 0

median in. mean.de

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610 people go on a coach trip each coach can hold 55 people how many coaches are needed
diamong [38]

Answer:

12

Step-by-step explanation:

610 divided by 55 = 11.09

but since u cant have 11.09, u need 12

4 0
2 years ago
Read 2 more answers
Each of 16 students measured the circumference of a tennis ball by four different methods, which were:Method A: Estimate the cir
lys-0071 [83]

Answer:

a) \bar X_A =22.744

\bar X_B =20.7

\bar X_C =21.013

\bar X_D =18.306

b) Median_A =\frac{23+24}{2}=23.5

Median_B =\frac{20.4+20.4}{2}=20.4

Median_C =\frac{21+21}{2}=21

Median_D =\frac{20.7+20.7}{2}=20.7

c) \bar X_A =23.25

\bar X_B =20.7

\bar X_C =21.04

\bar X_D =20.69

Step-by-step explanation:

Data given

Assuming the following data:

Method A: 18.0, 18.0, 18.0, 20.0, 22.0, 22.0, 22.5, 23.0, 24.0,24.0,25.0,25.0, 25.0,25.0,26.0,26.4

Method B: 18.8, 18.9, 18.9, 19.6, 20.1, 20.4, 20.4, 20.4, 20.4,20.5, 21.2. 22.0,22.0, 22.0,22.0,23.6

Method C: 20.2, 20.5, 20.5, 20.7, 20.8, 20.9, 21.0, 21.0,21.0,21.0, 21.0, 21.5,21.5,21.5,21.5,21.6

Method D: 20.0, 20.0, 20.0,-20.0, 20.2, 20.5, 20.5, 20.7, 20.7,20.7, 21.0, 21.1, 21.5, 21.6, 22.1,22.3.

Part a

The sample mean is defined as:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

If we apply this formula for the four methods we got:

\bar X_A =22.744

\bar X_B =20.7

\bar X_C =21.013

\bar X_D =18.306

Part b

Since the total number of points for each method is 16 and this is an even number we need to calculate the median as the average between the 8th and the 9th position of the data ordered from the smallest to the largest. If we do this we have that:

Median_A =\frac{23+24}{2}=23.5

Median_B =\frac{20.4+20.4}{2}=20.4

Median_C =\frac{21+21}{2}=21

Median_D =\frac{20.7+20.7}{2}=20.7

Part c

The trimmed mean by 20% means that we need to calculate removing the 20% from each of the tails, the 20% of 16 is 3.2, so we need to remove 3 observations from both ends of the data like this:

Method A: 20.0, 22.0, 22.0, 22.5, 23.0, 24.0,24.0,25.0,25.0, 25.0

Method B: 19.6, 20.1, 20.4, 20.4, 20.4, 20.4,20.5, 21.2. 22.0,22.0

Method C: 20.7, 20.8, 20.9, 21.0, 21.0,21.0,21.0, 21.0, 21.5,21.5

Method D: 20.0 20.2, 20.5, 20.5, 20.7, 20.7,20.7, 21.0, 21.1, 21.5.

If we calculate the mean for each method we got:

\bar X_A =23.25

\bar X_B =20.7

\bar X_C =21.04

\bar X_D =20.69

3 0
1 year ago
Roger bowled 7 games last weekend. His scores are 155, 165, 138, 172, 127, 193 , 142. What is the RANGE of Roger's scores?
Verdich [7]

Answer:

66

Step-by-step explanation:

In statistics, the formula for RANGE is given as the difference between the Highest and the Lowest value.

In the above values we are given data consisting of the 7 games that Roger bowled.

155, 165, 138, 172, 127, 193 , 142.

Step 1

We arrange from the least to the highest.

127, 138, 142, 155, 165, 172, 193

Step 2

Lowest value = 127

Highest value = 193

Step 3

Range = 193 - 127

= 66

Therefore, the range of Roger's scores is 66

4 0
2 years ago
To increase an amount by 7% what single multiplier would you use?
vazorg [7]
You would multiply by 1.7
8 0
2 years ago
Read 2 more answers
Which hyperbola has one focus point in common with the hyperbola (y+11)^2/(15^2)-(x-7)^2/(8^2)=1
irina [24]
\bf \textit{hyperbolas, vertical traverse axis }\\\\
\cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1
\qquad 
\begin{cases}
center\ ({{ h}},{{ k}})\\
vertices\ ({{ h}}, {{ k}}\pm a)
\end{cases}\\\\
-------------------------------\\\\
\cfrac{(y+11)^2}{15^2}-\cfrac{(x-7)^2}{8^2}=1\implies \cfrac{(y-(-11))^2}{15^2}-\cfrac{(x-7)^2}{8^2}=1\\\\
-------------------------------\\\\
c=\textit{distance from the center to either foci}\\\\
c=\sqrt{a^2+b^2}\implies c=\sqrt{15^2+8^2}\implies \boxed{c=17}

now, if you notice, the positive fraction, is the one with the "y" variable, and that simply means, the hyperbola traverse axis is over the y-axis, so it more or less looks like the picture below.

now, from the hyperbola form, we can see the center is at (7, -11), and that the "c" distance is 17.

so, from -11 over the y-axis, we move Up and Down 17 units to get the foci, which will put them at (7, -11-17) or (7, -28) and (7 , -11+17) or (7, 6)

5 0
2 years ago
Read 2 more answers
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