Answer:
Step-by-step explanation:
Since the line goes through the origin and is propotinal, we just do y/x
4/3= 1 1/3
y=1 1/3x
CHECK (Using desmos)
floyd caught a fish that weighed 2/3 pound. Kira caught a fish that weighed 7/8 pound. Whose fish weighed more? Explain the stra
2 answers:
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Answer:
El área es 2 veces el cuadrado del ancho del rectángulo.
Step-by-step explanation:
El área de un rectángulo viene dado por:
En donde:
a: es el ancho
l: es el largo
Si el largo es el doble del ancho:
Entonces el área es:
Por lo tanto, el área es 2 veces el cuadrado del ancho del rectángulo.
Espero que te sea de utilidad!
Answer:
y = 6
Step-by-step explanation:
y^2 + 36y - 252 = 0
(y-6)(y+42) = 0
y = 6 (accettable)
y = -42 (not accettable)
Answer:
A) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the first quartile.
Correct we satisfy that 1.2 <1.7 the lower limit
Step-by-step explanation:
Assuming this complete question: A violin student records the number of hours she spends practicing during each of nine consecutive weeks:
6.2 5.0 4.3 7.4 5.8 7.2 8.4 1.2 6.3
For this case we need to sort the data first on increasing way and we got:
1.2, 4.3, 5.0, 5.8, 6.2, 6.3, 7.2, 7.4, 8.4
For this case we have 9 values the median would be on the 5 position:
The first quartile would be 5 since we analyze 1.2, 4.3, 5.0, 5.8, 6.2 and the middle point is 5.0
The third quartile would be 7.2 since we analyze 6.2, 6.3, 7.2, 7.4, 8.4 and the middle point is 7.2
Then the interquartile rnage would be:
And
The lower limit on this case would be:
And our value is 1.2 is lower than the lower limit 1.7
Considering the smallest data value (1.2) and using the 1.5 × IQR rule, we would:
A) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the first quartile.
Correct we satisfy that 1.2 <1.7 the lower limit
B) not classify the value 1.2 as an outlier, because it is not more than 1.5 × IQR below the first quartile.
False 1.2 is more than 1.5 IQR below the first quartile
C) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the median.
False, we analyze if is an outlier comparing with the Q1 or Q3 not with the median
D) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the mean.
False we analyze if is an outlier comparing with the Q1 or Q3 not with the mean