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igor_vitrenko [27]
3 years ago
7

Simplify the expression. 55−4⋅5+12

Mathematics
2 answers:
coldgirl [10]3 years ago
5 0
First we add four point five plus twelve and we get sixteen point five, and so now we subtract fifty-five minus sixteen point five and we get our answer.

Answer: 62.5
Lunna [17]3 years ago
4 0
Your answer is 62.5. Hope i helped
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Please help rn thanks
Mama L [17]

9514 1404 393

Answer:

  (b)  (1, 1/6)

Step-by-step explanation:

The "unit rate" is always the y-value for x=1 when the relationship is proportional (as it is here).

That point is not specifically shown on the graph. It would be (1, 1/6).

8 0
2 years ago
The area of a trapezium is 800 cm2
svlad2 [7]

Answer:

The distance between the parallel sides is 20cm.

Step-by-step explanation:

Area of a trapezoid:

A=\frac{a+b}{2}\times h

where a and b are the 2 bases, and h is the height. The value we're trying to find here is the height.

3 of the 4 variables in this equation are already given:

A = 800

a = 48

b = 32

Just plug those in and solve for h:

\rightarrow 800=\frac{48+32}{2}\times h\\\rightarrow 800=\frac{80}{2}\times h\\\rightarrow 800=40\times h\\\rightarrow 800\div40=h\\\rightarrow h=20

5 0
2 years ago
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Mumz [18]
Can you take a screenshot
5 0
2 years ago
Find the original price given the total amount and tax or tip rate. Total price: $307.32 Tax or tip rate: 20%
dsp73
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307.32 - 61.46 = $245.86

8 0
3 years ago
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The box plots below show the distribution of salaries, in thousands, among employees of two small companies.
Shtirlitz [24]

Answer:

Mean and IQR

Step-by-step explanation:

The measure of centre gives the central or the measure which gives the best mid term of a distribution. Based in the details of the box plot, the median is the value which divides the box in the box plot.

For company A:

Range = 25 to 80 with a median value at 30 ; this means the median does not give a good centre measure of the distribution ad it is very close to the minimum value. This goes for the Company B plot too; with values ranging from (35 to 90) and the median designated at 40.

Hence, the mean will be the best measure of centre rather Than the median in this case.

For the variability, the interquartile range would best suit the distribution. With the lower quartile and upper quartile both having reasonable width to the minimum and maximum value of the distribution.

7 0
2 years ago
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