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hammer [34]
4 years ago
14

Which ordered pairs are solutions to the inequality x+3y≥−8?

Mathematics
1 answer:
Dima020 [189]4 years ago
6 0
I think the answer would be b
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There is 35 boys and 42 girls what is the ratio
8_murik_8 [283]
35:42, thats pretty much there all is to it... or 35/42
5 0
3 years ago
Read 2 more answers
Last week Isaiah and his roommates ate 3/8 of a carton of yogurt, and this week they ate
Sergio [31]

Answer: 5/24

Step-by-step explanation:

The problem is 7/12-3/8. I got a common denominator of 24, then multipled 8 and 3 by 3, and 7 and 12 by 2. The problem is now 14/25-9/24, which equals 5/24

6 0
3 years ago
Here are yesterday's high temperatures (in Fahrenheit) in 12 U.S. cities. 48, 50, 54, 56, 63, 63, 64, 68, 74, 74, 79, 80 Notice
irinina [24]

For the given data set

Minimum = 48

Lower quartile = 55

Median = 63.5

Upper quartile = 74

Maximum = 80

Interquartile range = 19

<h3>Measures of a Data </h3>

From the question, we are to determine the minimum, lower quartile, median, upper quartile, maximum, and interquartile range of the given data set

The given data set is

48, 50, 54, 56, 63, 63, 64, 68, 74, 74, 79, 80

Minimum = 48

Lower quartile = (54+56)/2

Lower quartile = 110/2

Lower quartile = 55

Median = (63+64)/2

Median = 127/2

Median = 63.5

Upper quartile = (74+74)/2

Upper quartile = 148/2

Upper quartile = 74

Maximum = 80

Interquartile range = Upper quartile - Lower quartile

Interquartile range = 74 - 55

Interquartile range = 19

Hence, for the given data set

Minimum = 48

Lower quartile = 55

Median = 63.5

Upper quartile = 74

Maximum = 80

Interquartile range = 19

Learn more on Measures of a Data here: brainly.com/question/15097997

#SPJ1

5 0
2 years ago
Use the quadratic formula to solve the equation <br> 25x^2-30x+25=0
Reika [66]

Answer: x=\frac{3}{5}+i\frac{4}{5},\:x=\frac{3}{5}-i\frac{4}{5}

Step-by-step explanation:

25x^2-30x+25=0

x_{1,\:2}=\frac{-\left(-30\right)\pm \sqrt{\left(-30\right)^2-4\cdot \:25\cdot \:25}}{2\cdot \:25}

\sqrt{\left(-30\right)^2-4\cdot \:25\cdot \:25}

=\sqrt{30^2-4\cdot \:25\cdot \:25}

=\sqrt{30^2-2500}

=i\sqrt{2500-30^2}

=40i

x_{1,\:2}=\frac{-\left(-30\right)\pm \:40i}{2\cdot \:25}

x_1=\frac{-\left(-30\right)+40i}{2\cdot \:25},\:x_2=\frac{-\left(-30\right)-40i}{2\cdot \:25}

x=\frac{3}{5}+i\frac{4}{5},\:x=\frac{3}{5}-i\frac{4}{5}

8 0
3 years ago
Don't answer this please Brodies
Nitella [24]
Okeyuyyyyyyyyyyyyy but why did you post it then
8 0
3 years ago
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