An increase on down payment, will decrease a mortgage payment
Answer
D. Down payment
We need to find the area under the standard normal curve that is to the right of 66.
Various ways of doing this include the following:
1) Calculator: normalcdf(66, 10000, 100, 15) = 0.988, or 98.8%
2) z-scores (using a published table of z scores)
66 - 100
a) Find the z score corresponding to 66. It is z = -------------- = -2.27
15
b) Find the area under the st. normal curve to the left of z=-2.27 (it is 0.012). Subtract that area from the total area, 1.000. The result will be the answer to this question.
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!
How many 2/3-quart portions are in 7 1/2 quarts of soup?
8
Answer:
18
Step-by-step explanation:
One way to solve this would be to just solve for random lengths, left to right, until we come to find UC.
We know JK = JH + HM + MK = 82 and JH = 22, so
82 = 22 + HM + MK
subtract 22 from both sides to isolate the unknowns
60 = HM + MK = HK
96 = HK + KU - HU
We know HK = 60
96 = 60 + KU
subtract 60 from both sides to isolate the unknown
We know KU = 36
105 = KN = KU + UC + CN
We know KU = 36 and CN = 51
105 = 36 + 51 + UC
105 = 87 + UC
subtract 87 from both sides to isolate the unknown
18 = UC
UC is what we're looking for, so the problem is solved
