Given:
<span>11 11.5 10.5 17 14.5 14.5 18 17 19
Arrange in chronological order from least to greatest.
10.5 ; 11 ; 11.5 ; 14.5 ; 14.5 ; 17 ; 17 ; 18 ; 19
</span><span>I used an online lower and upper fence calculator to get the necessary data.
Minimum: 10.5
Maximum: 19
Q1: 11.25
Q2 or median: 14.5
Q3: 17.5
Interquartile range can be solved by subtracting the value of Q1 from the value of Q3
IQR = Q3 - Q1
IQR = 17.5 - 11.25
IQR = 6.25 CHOICE A. </span>
Answer:
Sue would have taken 15 hours to sew all the costumes.
Step-by-step explanation:
In 10 hours, Sue sewed 5 costumes (1 every 2 hours).
Sue= 5 costumes
Sue sewed in 10 hours, twice as many costumes as Anne in 16 hours. Then 5 is twice costumes than Anne produced. Therefore, Anne sewed 2.5 costumes.
Anne: 2.5 costumes
In total they had to sew 5+2.5= 7.5 costumes. Then Sue would have taken 7.5*2= 15 hours to sew them all.
The answer is false. if the quantities combine to make zero
Answer:
B
Step-by-step explanation:
<span>It is estimated that 5000 live bacteria existed in the sample before treatment. After each day oftreatment, 40% of the sample remains alive. Which best describes the graph of the function thatrepresents the number of live bacteria after x days of treatment? f(x) = 5000(0.4)x, with a horizontal asymptote of y = 0 f(x) = 5000(0.6)x ...</span><span>
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