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Rashid [163]
3 years ago
9

Ill give correct one brainliest

Mathematics
1 answer:
adelina 88 [10]3 years ago
4 0

Answer: make 6 jumps which are 0.4 long to get 2.4

Step-by-step explanation:

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Determine the equation of the graph, and select the correct answer below.
Hunter-Best [27]
Vertex is (h,k)
Equation is ..
Y = (x -h)^2 + k
Put in vertex (-4,-1) gives the equation
Y = (x+4)^2 - 1
8 0
4 years ago
Please help asap <br>Thank you​
vodka [1.7K]

Answer:

k=70

Step-by-step explanation:

two ways to do this

the sum of k+110 has to equal 180, so k=70

52+58+k has to equal 180, so k=70

8 0
3 years ago
Read 2 more answers
Multiply each of the following numbers 7.32, and 0.006 by 10, 100, and 1,000. Then explain the pattern you can use to find the p
dangina [55]

7.32

10 = 73.2

100 = 732

1000 = 7320

0.006

10 = 0.06

100 = 0.6

1000 = 6

The decimal place moves to the right depending on how many zeros the number has.

For example, if you had 0.6 multiplied by 100, you would move 2 decimal places to the right because 100 has two zeros.

3 0
3 years ago
I NEED HELP AND IM ON A TIMED TEST
stira [4]

This Question is not Complete

Complete question

The number of absences per student in each of two classes is shown on the dot plots below.

A dot plot titled Number of absences per student in Misses Anderson's class. The number line goes from 0 to 6. There is 1 dot above 0, 3 above 1, 3 above 2, 3 above 3, 1 above 4, and 0 above 5 and 6.

A dot plot titled Number of absences per student in Misses Bergot's class. The number line goes from 0 to 6. There is 1 dot above 0, 1 above 1, 2 above 2, 3 above 3, 2 above 4, 1 above 5, and 1 above 6.

Which statements accurately compare the two data sets? Select three options.

a. The mean number of absences is greater in Mrs. Anderson’s class.

b. The mean number of absences is greater in Mrs. Bergot’s class.

c. There is more variability in Mrs. Anderson’s data set.

d. There is more variability in Mrs. Bergot’s data set.

e. The mean and MAD are better representations of these data sets than the median and IQR.

Answer:

b. The mean number of absences is greater in Mrs. Bergot’s class.

d. There is more variability in Mrs. Bergot’s data set.

e. The mean and MAD are better representations of these data sets than the median and IQR.

Step-by-step explanation:

From the above data given in the question and the options given to us, we can observe from the comparison of both the number of absences in Mrs Bergot's class to that in Mrs Anderson's class, there will be a presence of variation in the data of one of the classes and that class happens to be Mrs Bergot's class.

The appropriate statistical tool that we can be used to determine and measure the centre of variability of the data is Mean and Mean Absolute Deviation.

7 0
3 years ago
I don't know how to do these sort of problems... Can someone help me out?
VLD [36.1K]
180=76+4x
180-76=4x
104=4x
104/4=x
26=x
Measurement A= 26
6 0
3 years ago
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