C. Action: Divide both sides by 3.
<span> 7/4 = 1.75000 is the other fraction</span>
Answer:
73.1 inches
Step-by-step explanation:
We solve for this using z score formula.
z-score is z = (x-μ)/σ,
where x is the raw score,
μ is the population mean
σ is the population standard deviation.
x = unknown = randomly selected male's height
μ = 70.3 inches
σ = 4.1 inches
z = is not given but we are given its percentile which is 75
The z score for a 75th percentile = 0.674
Hence,
0.674 = x - 70.3/4.1
Cross Multiply
0.674 × 4.1 = x - 70.3
2.7634 = x - 70.3
x = 70.3 + 2.7634
x = 73.0634 inches
The height of the randomly selected male = 73.0634
Therefore, the height of the randomly selected male to the nearest tenth of an inch is 73.1 inches
Answer:
Kela will be able to travel for 7 stops before her money runs out.
Step-by-step explanation:
Since Kela wants to visit a friend who lives 8 kilometers away, and she'll ride the subway as she can before walking the rest of the way, and she needs to buy an access pass that costs $ 5.50 and there is also a fee of $ 1.25 for each stop, if Kela doesn't want to spend more than $ 15 on the trip, to determine the largest number of stops she can afford, the following calculation must be performed:
(15 - 5.5) / 1.25 = X
9.5 / 1.25 = X
7.6 = X
Therefore, Kela will be able to travel for 7 stops before her money runs out.
Answer:
IQR = 6
Step-by-step explanation:
Given:
The given set of data is:
40 41 42 43 44 45 46 47 48 49 50
Number of terms, 
Therefore, the median is given as
term which is the sixth term. Median = 45
(40 41 42 43 44) <u>45</u> (46 47 48 49 50 )
Now, we find the median of the lower half (
) and upper half (
)of the data that are before and after the median.
The lower half of the data has the values 40 41 <u>42</u> 43 44.
Number of terms are 5. So, median is the third term which is, 
The upper half of the data has the values 46 47 <u>48</u> 49 50.
Number of terms are 5. So, median is the third term which is, 
Now, IQR is given as the difference of upper half median and lower half median.
