The resultant of twelve times two plus the resultant of four times three.
Answer:
Step-by-step explanation:
In order to solve this problem, one must use the trigonometric ratios. These ratios are the following,
Remember, each side is named relative to the angle, thus sides will attain different names depending on the angle which one sues to calculate with.
In this case, one is given the measure of the angle, and the measure of the side opposite to the angle. One is asked to find the measure of the side adjacent to the angle. Use the trigonometric ratio (
) to solve for the unknown. Substitute in the given values,
Manipulate the equation such that it is solved for the parameter(x),
Solve,
Answer:
You should have $1,463.95
Step-by-step explanation:
The formula for this is A=P(1+r)^x
P being the initial amount, r being the percentage, and x being the time passing.
<u>1. Write out the equation:</u> A=1400(1+0.015)^3
REMEMBER- writing percentages into decimals is moving the decimal point two times to the left--or dividing by 100!
<u>2. Parentheses:</u> 1400(1.015)^3
At this point, you're going to want to do exponents first (PEMDAS). If you're typing this into a calculator it should just give you the answer right away with the full equation.
<u>3. Answer:</u> 1463.95 (rounded to the nearest hundredth).
**I apologize if this is wrong; I haven't done it in a while so I don't know if I did 3 years or 3 months.. lol. That's probably not helping with the confidence level, but good luck!
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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!
The answer is 480 because you need to multiply the months by the cost 40(12) which is 480.
