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laiz [17]
3 years ago
5

Which graphic organizer is most helpful when comparing and contrasting information?

Mathematics
2 answers:
UNO [17]3 years ago
7 0

Answer:

The answer is the Venn diagram.

Step-by-step explanation:

The graphic organizer that is most helpful when comparing and contrasting information is the Venn diagram.

The Venn diagram can compare and contrast with the help of circles drawn to represent different sets.

The middle overlapping area of circles is used for denoting similarities between sets.

Musya8 [376]3 years ago
5 0
A. a Venn diagram. A Venn diagram is two circles that overlap in the middle. Each circle represents a different subject. The overlapping part is used for comparing, while the rest of the circles are used for contrasting.
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Deanna has $150.00 in her account.At the end of each week, she plans to take $15.00 out of her account for her spending money. W
AlexFokin [52]

Let the balance = Y and the number of weeks = x

She takes out 15 per week, so multiply 15 by x ( the number of weeks) to get 15x.

You would then want to subtract that from the amount she started with in her account.

The equation becomes: Y = 150 - 15x

3 0
3 years ago
Identify the zero(s) of the function graphed below
k0ka [10]

Answer:

Step-by-step explanation:

The zeros are the values of x for which y=0.

The zero of this graph is x=1.

8 0
2 years ago
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ankoles [38]
Go to your profile and you should see it
3 0
3 years ago
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By how much does 5 exceed -4?<br>(a) 1<br>(b)-1<br>(c) 9<br>(d) -9​
Elden [556K]

Answer:

C

Step-by-step explanation:

-4+9=5

6 0
3 years ago
The lifespans of lions in a particular zoo are normally distributed. The average lion lives 12.5 years; the standard
bulgar [2K]

We have been given that the lifespans of lions in a particular zoo are normally distributed. The average lion lives 12.5 years; the standard  deviation is 2.4 years. We are asked to find the probability of a lion living longer than 10.1 years using empirical rule.

First of all, we will find the z-score corresponding to sample score 10.1.

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

z = z-score,

x = Random sample score,

\mu = Mean

\sigma = Standard deviation.

z=\frac{10.1-12.5}{2.4}

z=\frac{-2.4}{2.4}

z=-1

Since z-score of 10.1 is -1. Now we need to find area under curve that is below one standard deviation from mean.

We know that approximately 68% of data points lie between one standard deviation from mean.

We also know that 50% of data points are above mean and 50% of data points are below mean.

To find the probability of a data point with z-score -1, we will subtract half of 68% from 50%.

\frac{68\%}{2}=34\%

50\%-34\%=16\%

Therefore, the probability of a lion living longer than 10.1 years is approximately 16%.

3 0
3 years ago
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