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sladkih [1.3K]
3 years ago
13

Is 5 in the solution set of x+3>8

Mathematics
2 answers:
Irina-Kira [14]3 years ago
8 0

Answer:

It 5 would not be x

Step-by-step explanation:

If 5 was x, then it would be equivalent, because 5+3 = 8. I hope this helped.

vfiekz [6]3 years ago
8 0
It would be greater than 5 so

6 and above
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Answer:

B. 24.3°

Step-by-step explanation:

Apply the sine rule to solve for x

Thus:

\frac{sin(x)}{8} = \frac{sin(112)}{18}

\frac{sin(x)}{8} = \frac{sin(112)}{18}

Multiply both sides by 8

sin(x)= \frac{sin(112)}{18}*8

sin(x) = 0.4121

x = sin^{-1}(0.4121}

x = 24.3° (nearest tenth)

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3 years ago
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6 0
3 years ago
For which operations is the set {–1, 0, 1} closed? Choose all answers that are correct. A. addition B. division C. multiplicatio
spayn [35]
Hello,

Addition : no because -1+(-1)=-2 not in set.
Division : no because 1/0 is not in the set :infinity
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5 0
3 years ago
The table shows the heights of 40 students in a class.
Studentka2010 [4]

Answer:

Step-by-step explanation:

Hello!

a)

The given information is displayed in a frequency table, since the variable of interest "height of a student" is a continuous quantitative variable the possible values of height are arranged in class intervals.

To calculate the mean for data organized in this type of table you have to use the following formula:

X[bar]= (∑x'fi)/n

Where

x' represents the class mark of each class interval and is calculated as (Upper bond + Lower bond)/2

fi represents the observed frequency for each class

n is the total of observations, you can calculate it as ∑fi

<u>Class marks:</u>

x₁'= (120+124)/2= 122

x₂'= (124+128)/2= 126

x₃'= (128+132)/2= 130

x₄'= (132+136)/2= 134

x₅'= (136+140)/2= 138

Note: all class marks are always within the bonds of its class interval, and their difference is equal to the amplitude of the intervals.

n= 7 + 8 + 13 + 9 + 3= 40

X[bar]= (∑x'fi)/n= [(x₁'*f₁)+(x₂'*f₂)+(x₃'*f₃)+(x₄'*f₄)+(x₅'*f₅)]/n) = [(122*7)+(126*8)+(130*13)+(134*9)+(138*3)]/40= 129.3

The estimated average height is 129.3cm

b)

This average value is estimated because it wasn't calculated using the exact data measured from the 40 students.

The measurements are arranged in class intervals, so you know, for example, that 7 of the students measured sized between 120 and 124 cm (and so on with the rest of the intervals), but you do not know what values those measurements and thus estimated a  mean value within the interval to calculate the mean of the sample.

I hope this helps!

4 0
4 years ago
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