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2 years ago

The histogram shows the heights in means of trees in a certain section of a park

Mathematics

1 answer:

aleksley [76]2 years ago

6 0

<h2>Answer:</h2>

The number of trees which are less than 20 meters are:

<h2>Step-by-step explanation:</h2>

Based on the histogram we are asked to find the number of trees which are less than 20 meters in height.

Class interval Frequency

4-8 1

8-12 2

12-16 2

16-20 4

20-24 6

24-28 4

28-32 2

i.e. we are asked to find the total number of tress that lie in the class interval 4-20.

i.e. we add the frequencies that lie in the interval 4-20

i.e. Number of trees= 1+2+2+4=9 trees.

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Hope researches the impact of one variable on another. She graphs the data from her research and sees that the data forms a shap

zaharov [31]

Answer:

She can determine that while there is an association between the variables, there is no correlation, and she cannot determine causation.

Step-by-step explanation:

Since there is an arch shape to her graph, we know that as one variable changes, the other changes in the same manner. This means there is an association between the variables.

However, since the graph is not linear, there is no correlation between the variables. Since there is no correlation, we cannot determine causation.

Mark me as brainiest

6 0

2 years ago

Read 2 more answers

Anyone know the answer

Arlecino [84]

Answer: A: 2

Step-by-step explanation:

In order to find the slope of the line, you must know about rise/run, which means that you go up and over to your next point. For example: I started at the point 0,-4 and went up two and to the right one to the point -2,1, which i rose 2 and ran 1, or 2/1, which equals 2

3 0

2 years ago

If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?

Alisiya [41]

If $k$ is odd, then

$\displaystyle\sum_{n=1}^{\lfloor k/2\rfloor}2n=2\dfrac{\left\lfloor\frac k2\right\rfloor\left(\left\lfloor\frac k2\right\rfloor+1\right)}2=\left\lfloor\dfrac k2\right\rfloor^2+\left\lfloor\dfrac k2\right\rfloor$

while if $k$ is even, then the sum would be

$\displaystyle\sum_{n=1}^{k/2}2n=2\dfrac{\frac k2\left(\frac k2+1\right)}2=\dfrac{k^2+2k}4$

The latter case is easier to solve:

$\dfrac{k^2+2k}4=2k\implies k^2-6k=k(k-6)=0$

which means $k=6$ .

In the odd case, instead of considering the above equation we can consider the partial sums. If $k$ is odd, then the sum of the even integers between 1 and $k$ would be

$S=2+4+6+\cdots+(k-5)+(k-3)+(k-1)$

Now consider the partial sum up to the second-to-last term,

$S^*=2+4+6+\cdots+(k-5)+(k-3)$

Subtracting this from the previous partial sum, we have

$S-S^*=k-1$

We're given that the sums must add to $2k$ , which means

$S=2k$
$S^*=2(k-2)$

But taking the differences now yields

$S-S^*=2k-2(k-2)=4$

and there is only one $k$ for which $k-1=4$ ; namely, $k=5$ . However, the sum of the even integers between 1 and 5 is $2+4=6$ , whereas $2k=10\neq6$ . So there are no solutions to this over the odd integers.

5 0

2 years ago

Is there anybody taking the algebra test in texas i would really appreciate the answeres

s344n2d4d5 [400]

Answer:

which one, i just took a test, and im from texas

Step-by-step explanation:

6 0

2 years ago

Four horses ran in a race

irina [24]

Answer:

Dee won

Step-by-step explanation:

Using < to mean "finished ahead of ", we are given ...

B = 3

D < B

D < C

A < C

D < A

Dee finished ahead of Air, Bart, and Caper, so Dee won the race.

_____

The order of finish must be D, A, B, C.

3 0

3 years ago