Answer: The percentage change is 56.67%.
Step-by-step explanation:
From 120 meals to 52 meals, change in meals = ( 120- 52) meals
= 68 meals
The percentage change = [tex]\dfrac{\text{change in meals}}{\text{Original quantity of meals}}\times100[/tex]
[tex]=\dfrac{68}{120}\times100\\\\=56.67\%[/tex]
Hence, the percentage change is 56.67%.
Find usubscript10 in the sequence -23, -18, -13, -8, -3, ...
Step-by-step explanation:
utilise the formula a+(n-1)d
a is the first number while d is common difference
Answer:
22
Step-by-step explanation:
Using the formular, Un = a + (n - 1)d
Where n = 10; a = -23; d = 5
U10 = -23 + (9)* 5
U10 = -23 + 45 = 22
The cost, C, in United States Dollars ($), of cleaning up x percent of an oil spill along the Gulf Coast of the United States increases tremendously as x approaches 100. One equation for determining the cost (in millions $) is:
Complete Question
On the uploaded image is a similar question that will explain the given question
Answer:
The value of k is [tex]k = 214285.7[/tex]
The percentage of the oil that will be cleaned is [tex]x = 80.77\%[/tex]
Step-by-step explanation:
From the question we are told that
The cost of cleaning up the spillage is [tex]C = \frac{ k x }{100 - x }[/tex] [tex]x \le x \le 100[/tex]
The cost of cleaning x = 70% of the oil is [tex]C = \$500,000[/tex]
Now at [tex]C = \$500,000[/tex] we have
[tex]\$ 500000 = \frac{ k * 70 }{100 - 70 }[/tex]
[tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]
[tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]
[tex]k = 214285.7[/tex]
Now When [tex]C = \$900,000[/tex]
[tex]x = 80.77\%[/tex]
which of the following not between -10 and -8
-17/2
-7
-9
-8.5
Answer:
-7Step-by-step explanation:
This is best read on the number line.
Look at the picture.
[tex]-\dfrac{17}{2}=-8\dfrac{1}{2}=-8.5[/tex]
cooks are needed to prepare for a large party. Each cook can bake either 5 Large cakes or 14 small cakes per hour . The kitchen is available for 3 hours and 29 large cakes and 260 cakes need to be baked . How many cooks are required to bake the required number of cakes during the time the kitchen is available?
it was all about equating some values
to bake the required number of cakes during the available 3-hour time period, 7 cooks are required.
Let's determine the number of cooks required to bake the required number of cakes during the available time.
We have the following information:
- Each cook can bake either 5 large cakes or 14 small cakes per hour.
- The kitchen is available for 3 hours.
- We need to bake 29 large cakes and 260 cakes in total.
First, let's calculate the number of large cakes that can be baked by one cook in 3 hours:
1 cook can bake 5 large cakes/hour × 3 hours = 15 large cakes.
Next, let's calculate the number of small cakes that can be baked by one cook in 3 hours:
1 cook can bake 14 small cakes/hour × 3 hours = 42 small cakes.
Now, let's calculate the number of large cakes that can be baked by all the cooks in 3 hours:
Total number of large cakes = Number of cooks × Large cakes per cook per 3 hours
We need to bake 29 large cakes, so:
29 = Number of cooks × 15
Number of cooks = 29 / 15 ≈ 1.93
Since we can't have a fraction of a cook, we need to round up to the nearest whole number. Therefore, we need at least 2 cooks to bake the required number of large cakes.
Similarly, let's calculate the number of small cakes that can be baked by all the cooks in 3 hours:
Total number of small cakes = Number of cooks × Small cakes per cook per 3 hours
We need to bake 260 small cakes, so:
260 = Number of cooks × 42
Number of cooks = 260 / 42 ≈ 6.19
Again, rounding up to the nearest whole number, we need at least 7 cooks to bake the required number of small cakes.
Since we need to satisfy both requirements for large and small cakes, we choose the larger number of cooks required, which is 7 cooks.
Therefore, to bake the required number of cakes during the available 3-hour time period, 7 cooks are required.
Learn more about work here
https://brainly.com/question/13245573
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Which of the following represents "next integer after the integer n"? n + 1 n 2n
Answer:
n + 1
Step-by-step explanation:
Starting with the integer 'n,' we represent the "next integer" by n + 1.
Suppose that a sample mean is .29 with a lower bound of a confidence interval of .24. What is the upper bound of the confidence interval?
Answer:
The upper bound of the confidence interval is 0.34
Step-by-step explanation:
Here in this question, we want to calculate the upper bound of the confidence interval.
We start by calculating the margin of error.
Mathematically, the margin of error = 0.29 -0.24 = 0.05
So to get the upper bound of the confidence interval, we simply add this margin of error to the mean
That would be 0.05 + 0.29 = 0.34
In a binomial distribution, n = 8 and π=0.36. Find the probabilities of the following events. (Round your answers to 4 decimal places.)
a. x=5
b. x <= 5
c. x>=6
Answer:
[tex]\mathbf{P(X=5) =0.0888}[/tex]
P(x ≤ 5 ) = 0.9707
P ( x ≥ 6) = 0.0293
Step-by-step explanation:
The probability of a binomial mass distribution can be expressed with the formula:
[tex]\mathtt{P(X=x) =(^{n}_{x} ) \ \pi^x \ (1-\pi)^{n-x}}[/tex]
[tex]\mathtt{P(X=x) =(\dfrac{n!}{x!(n-x)!} ) \ \pi^x \ (1-\pi)^{n-x}}[/tex]
where;
n = 8 and π = 0.36
For x = 5
The probability [tex]\mathtt{P(X=5) =(\dfrac{8!}{5!(8-5)!} ) \ 0.36^5 \ (1-0.36)^{8-5}}[/tex]
[tex]\mathtt{P(X=5) =(\dfrac{8!}{5!(3)!} ) \ 0.36^5 \ (0.64)^{3}}[/tex]
[tex]\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 \times 5!}{5!(3)!} ) \times \ 0.0060466 \ \times 0.262144}[/tex]
[tex]\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 }{3 \times 2 \times 1} ) \times \ 0.0060466 \ \times 0.262144}[/tex]
[tex]\mathtt{P(X=5) =({8 \times 7 } ) \times \ 0.0060466 \ \times 0.262144}[/tex]
[tex]\mathtt{P(X=5) =0.0887645}[/tex]
[tex]\mathbf{P(X=5) =0.0888}[/tex] to 4 decimal places
b. x ≤ 5
The probability of P ( x ≤ 5)[tex]\mathtt{P(x \leq 5) = P(x = 0)+ P(x = 1)+ P(x = 2)+ P(x = 3)+ P(x = 4)+ P(x = 5})[/tex]
[tex]{P(x \leq 5) = ( \dfrac{8!}{0!(8!)} \times (0.36)^0 \times (1-0.36)^8 \ ) + \dfrac{8!}{1!(7!)} \times (0.36)^1 \times (1-0.36)^7 \ +[/tex][tex]\dfrac{8!}{2!(6!)} \times (0.36)^2 \times (1-0.36)^6 \ + \dfrac{8!}{3!(5!)} \times (0.36)^3 \times (1-0.36)^5 + \dfrac{8!}{4!(4!)} \times (0.36)^4 \times (1-0.36)^4 \ + \dfrac{8!}{5!(3!)} \times (0.36)^5 \times (1-0.36)^3 \ )[/tex]
P(x ≤ 5 ) = 0.0281+0.1267+0.2494+0.2805+0.1972+0.0888
P(x ≤ 5 ) = 0.9707
c. x ≥ 6
The probability of P ( x ≥ 6) = 1 - P( x ≤ 5 )
P ( x ≥ 6) = 1 - 0.9707
P ( x ≥ 6) = 0.0293
Simplify 5(R + 2) - 6.
5R + 4
5R - 4
5R - 6
Step-by-step explanation:
Hey, there!!
5(R+2)-6.
Fistly multiply (R+2) by 5.
=5R + 10 - 6
Subtract 6 from 10.
= 5R +4.
Therefore, 5R + 4 is correct answer.
{ While simplifying the expression if there is multiplication or divide do it first and then add or or subtract like terms to get the simplified form of the expressions. }
Hope it helps..
Gail paid a total of $12,000 for stock that was $6 per share. If she sold all her shares for $18,000, how much profit on each share did she make?
A
$9
B
$3
С.
S2000
D
$6.000
Answer:
$3
Step-by-step explanation:
Given
Total Cost Price: $12,000
Unit Cost Price= $6
Total Selling Price = $18,000
Required
Determine the profit on each share
First, we need to determine the units of share bought;
Units = Total cost price / Unit Cost Price
[tex]Units = \frac{\$12000}{\$6}[/tex]
[tex]Units = 2000[/tex]
Next is to determine the selling price of each share; This is calculated as follows;
Unit Selling Price = Total Selling Price / Units Sold
[tex]Unit\ Selling\ Price = \frac{\$18000}{\$2000}[/tex]
[tex]Unit\ Selling\ Price = \$9[/tex]
The profit is the difference between the unit cost price and unit selling price
[tex]Profit = Unit\ Selling\ Price - Unit\ Cost\ Price[/tex]
[tex]Profit = \$9 - \$6[/tex]
[tex]Profit = \$3[/tex]
Find a cubic polynomial with integer coefficients that has $\sqrt[3]{2} + \sqrt[3]{4}$ as a root.
Find the powers [tex]a=\sqrt{2}+\sqrt{3}[/tex]
$a^{2}=5+2 \sqrt{6}$
$a^{3}=11 \sqrt{2}+9 \sqrt{3}$
The cubic term gives us a clue, we can use a linear combination to eliminate the root 3 term $a^{3}-9 a=2 \sqrt{2}$ Square $\left(a^{3}-9 a\right)^{2}=8$ which gives one solution. Expand we have $a^{6}-18 a^{4}-81 a^{2}=8$ Hence the polynomial $x^{6}-18 x^{4}-81 x^{2}-8$ will have a as a solution.
Note this is not the simplest solution as $x^{6}-18 x^{4}-81 x^{2}-8=\left(x^{2}-8\right)\left(x^{4}-10 x^{2}+1\right)$
so fits with the other answers.
Answer:
[tex]y^3 -6y-6[/tex]
solve this equation 4log√x - log 3x =log x^2
Answer:
[tex]x = \frac{1}{3} [/tex]
Step-by-step explanation:
*Move terms to the left and set equal to zero:
4㏒(√x) - ㏒(3x) - ㏒(x²) = 0
*simplify each term:
㏒(x²) - ㏒(3x) - ㏒(x²)
㏒(x²÷x²) -㏒(3x)
㏒(x²÷x² / 3x)
*cancel common factor x²:
㏒([tex]\frac{1}{3x}[/tex])
*rewrite to solve for x :
10⁰ = [tex]\frac{1}{3x}[/tex]
1 = [tex]\frac{1}{3x}[/tex]
1 · x = [tex]\frac{1}{3x}[/tex] · x
1x = [tex]\frac{1}{3}[/tex]
*that would be our answer, however, the convention is to exclude the "1" in front of variables so we are left with:
x = [tex]\frac{1}{3}[/tex]
A bag of 100 hard candies included 30 butterscotch, 40 peppermint, 15 strawberry, 10 orange, and 5 banana. The probability that the first candy pulled out of the bag will be butterscotch or strawberry is .45
a) true
b) false
Answer:
true
Step-by-step explanation:
there is 100 candies. That means we can easily turn the amount of each type of candy into a percent. there was 30 butterscotch which means that is 30 percent. There was 15 strawberry which means that is 15 percent. add that and you get 45. This is a shortcut and i advise you use the way your teacher taught you.
[tex]|\Omega|=100\\|A|=30+15=45\\\\P(A)=\dfrac{45}{100}=0.45[/tex]
So TRUE
given the vector with a manitude of 9m at an angle a of -80 degrees, decompose this vector into two vector components oarallel to the x axis with a slope of
Answer:
We have the magnitude, M, and the angle A.
(The angle is always measured from the +x-axis)
Then we have that:
x = M*cos(A)
y = M*sin(A)
in this case:
M = 9m
A = -80°
x = 9m*cos(-80°) = 1.562
y = 9m*sin(-80) = -8.86m
Now, the component parallel to the x axis is:
x = 9m*cos(-80°) = 1.562 m
And the slope of something parallel to the x-axis is always zero, as this is a constant line.
Use Lagrange multipliers to find three numbers whose sum is 30 and the product P = x3y4z is a maximum. Choose the answer for the smallest of the three values. Question 20 options: a) 21/4 b) 5 c) 15/4 d) 3
We want to maximize [tex]x^3y^4z[/tex] subject to the constraint [tex]x+y+z=30[/tex].
The Lagrangian is
[tex]L(x,y,z,\lambda)=x^3y^4z-\lambda(x+y+z-30)[/tex]
with critical points where the derivatives vanish:
[tex]L_x=3x^2y^4z-\lambda=0[/tex]
[tex]L_y=4x^3y^3z-\lambda=0[/tex]
[tex]L_z=x^3y^4-\lambda=0[/tex]
[tex]L_\lambda=x+y+z-30=0[/tex]
[tex]\implies\lambda=3x^2y^4z=4x^3y^3z=x^3y^4[/tex]
We have
[tex]3x^2y^4z-4x^3y^3z=x^2y^3z(3y-4x)=0\implies\begin{cases}x=0,\text{ or}\\y=0,\text{ or}\\z=0,\text{ or}\\3y=4x\end{cases}[/tex]
[tex]3x^2y^4z-x^3y^4=x^2y^4(3z-x)=0\implies\begin{cases}x=0,\text{ or}\\y=0,\text{ or}\\3z=x\end{cases}[/tex]
[tex]4x^3y^3z-x^3y^4=x^3y^3(4z-y)=0\implies\begin{cases}x=0,\text{ or}\\y=0,\text{ or}4z=y\end{cases}[/tex]
Let's work with [tex]x=3z[/tex] and [tex]y=4z[/tex], for which we have
[tex]x+y+z=8z=30\implies z=\dfrac{15}4\implies\begin{cases}x=\frac{45}4\\y=15\end{cases}[/tex]
The smallest of these is C. 15/4.
Consider the function below. (If an answer does not exist, enter DNE.) f(x) = x3 − 27x + 3 (a) Find the interval of increase. (Enter your answer using interval notation.)
Answer:
(-∞,-3) and (3,∞)
Step-by-step explanation:
f(x) = x³ − 27x + 3
1. Find the critical points
(a) Calculate the first derivative of the function.
f'(x) = 3x² -27
(b) Factor the first derivative
f'(x)= 3(x² - 9) = 3(x + 3) (x - 3)
(c) Find the zeros
3(x + 3) (x - 3) = 0
x + 3 = 0 x - 3 = 0
x = -3 x = 3
The critical points are at x = -3 and x = 3.
2. Find the local extrema
(a) x = -3
f(x) = x³ − 27x + 3 = (-3)³ - 27(-3) + 3 = -27 +81 + 3 = 57
(b) x = 3
f(x) = x³ − 27x + 3 = 3³ - 27(3) + 3 = 27 - 81 + 3 = -51
The local extrema are at (-3,57) and (3,-51).
3, Identify the local extrema as maxima or minima
Test the first derivative (the slope) over the intervals (-∞, -3), (-3,3), (3,∞)
f'(-4) = 3x² -27 = 3(4)² - 27 = 21
f'(0) = 3(0)² -27 = -27
f'(4) = 3(4)² - 27 = 51
The function is increasing on the intervals (-∞,-3) and (3,∞).
The graph below shows the critical points of your function.
Find the principal invested if $495 interest was earned in 3 years at an interest rate of 6%.
Answer: $2750
Step-by-step explanation:
Formula to calculate interest : I = Prt , where P = Principal amount , r = rate of interest ( in decimal) , t= time.
Given: I= $495
t= 3 years
r= 6% = 0.06
Then, according to the above formula:
[tex]495 = P (0.06\times3)\\\\\Rightarrow\ P=\dfrac{495}{0.18}\\\\\Rightarrow\ P=2750[/tex]
Hence, the principal invested = $2750
You are investing $5,000 and can invest for 2 years or 3 years at 1.75% and 1.25% interest rates, respectively. Which earns more interest
Answer:
The 3 years investment earns more interest
Step-by-step explanation:
Given
Principal, P = $5,000
Required
Determine which earns more interest
When Rate = 1.75% and Year = 2
Interest is as follows;
[tex]I = \frac{PRT}{100}[/tex]
Substitute 1.75 for R, 5000 for P and 2 for T
[tex]I = \frac{5000 * 1.75 * 2}{100}[/tex]
[tex]I = \frac{17500}{100}[/tex]
[tex]I = \$175[/tex]
When Rate = 1.25% and Year = 3
Interest is as follows;
[tex]I = \frac{PRT}{100}[/tex]
Substitute 1.25 for R, 5000 for P and 3 for T
[tex]I = \frac{5000 * 1.25 * 3}{100}[/tex]
[tex]I = \frac{18750}{100}[/tex]
[tex]I = \$187.5[/tex]
Comparing the interest of both investments, the 3 years investment earns more interest
Layla is going to drive from her house to City A without stopping. Layla plans to drive
at a speed of 30 miles per hour and her house is 240 miles from City A. Write an
equation for D, in terms of t, representing Layla's distance from City A t hours after
leaving her house.
Answer:
D = 240 - 30t
Step-by-step explanation:
If the equation represents her distance from City A, we need to include 240 in the equation to represent the distance to the city.
Then, we need to subtract 30t from 240 in the equation because 30t represents how far she will have traveled in t hours.
So, D = 240 - 30t is the equation that will represent Layla's distance from the city.
What is the mulitplicative rate of change for the exponential function f(x) = 2 (5over2) to the negative x power ?
Answer:
2/5
Step-by-step explanation:
f(x) = 2(5/2)^-x = 2(2/5)^x
The multiplicative rate of change is the base of the positive exponent, 2/5.
A random sample of 149 recent donations at a certain blood bank reveals that 76 were type A blood. Does this suggest that the actual percentage of type A donations differs from 40%, the percentage of the population having type A blood? Carry out a test of appropriate hypotheses using a significance level of 0.01. Would your conclusion have been different if a significance level of 0.05 has been used?
Answer:
Yes it suggest that the actual percentage of type A donations differs from 40%, the percentage of the population having type A blood.
Well if a significance level of 0.05 is used it will not affect the conclusion
Step-by-step explanation:
From the question we are told that
The sample size is [tex]n = 149[/tex]
The number that where type A blood is k = 76
The population proportion is [tex]p = 0.40[/tex]
The significance level is [tex]\alpha = 0.01[/tex]
Generally the sample proportion is mathematically represented as
[tex]\r p = \frac{k}{n}[/tex]
=> [tex]\r p = \frac{76}{149}[/tex]
=> [tex]\r p = 0.51[/tex]
The Null hypothesis is [tex]H_o : p = 0.41[/tex]
The Alternative hypothesis is [tex]H_a : p \ne 0.40[/tex]
Next we obtain the critical value of [tex]\alpha[/tex] from the z-table.The value is
[tex]Z_{\alpha } = Z_{0.01} = 1.28[/tex]
Generally the test statistics is mathematically evaluated as
[tex]t = \frac{\r p - p }{ \sqrt{ \frac{p(1-p)}{n} } }[/tex]
substituting values
[tex]t = \frac{0.51 - 0.40 }{ \sqrt{ \frac{0.40 (1-0.40 )}{149} } }[/tex]
[tex]t =2.74[/tex]
So looking at the values for t and [tex]Z_{0.01}[/tex] we see that [tex]t > Z_{0.01}[/tex] so we reject the null hypothesis. Which means that there is no sufficient evidence to support the claim
Now if [tex]\alpha = 0.05[/tex] , the from the z-table the critical value for [tex]\alpha = 0.05[/tex] is [tex]Z_{0.05} = 1.645[/tex]
So comparing the value of t and [tex]Z_{0.05} = 1.645[/tex] we see that [tex]t > Z_{0.05}[/tex] hence the conclusion would not be different.
Identify the decimals labeled with the letters A, B, and C on the scale below. Letter A represents the decimal Letter B represents the decimal Letter C represents the decimal
[tex]10[/tex] divisions between $389$ and $390$ so each division is $\frac{390-389}{10}=0.1$
A is 8 division from $389$, so, A is $389+8\times 0.1=389.8$
similarly, C is one division behind $389$ so it is $389-1\times 0.1=388.9$
and B is $390.3$
How do you compress this?
[tex]\displaystyle\\(a+b)^n\\T_{r+1}=\binom{n}{r}a^{n-r}b^r\\\\\\(x+2)^7\\a=2x\\b=3\\r+1=4\Rightarrow r=3\\n=5\\T_4=\binom{5}{3}\cdot (2x)^{5-3}\cdot3^3\\T_4=\dfrac{5!}{3!2!}\cdot 4x^2\cdot27\\T_4=\dfrac{4\cdot5}{2}\cdot 4x^2\cdot27\\\\T_4=1080x^2[/tex]
logx-log(x-l)^2=2log(x-1)
Answer:
x = 1.00995066776
x = 2.52925492433
Step-by-step explanation:
This sort of equation is best solved using a graphing calculator. For that purpose, I like to rewrite the equation as a function whose zeros we're seeking. Here, that becomes ...
[tex]f(x)=\log{(x)}-\log{(x-1)}^2-2\log{(x-1)}[/tex]
The attached graph shows zeros at
x = 1.00995066776 and 2.52925492433
_____
Comment on the equation
Note that we have taken the middle term to be the square of the log, rather than the log of a square. For the latter interpretation, see mberisso's answer at https://brainly.com/question/17210068
Comment on the answer refinement
We have used Newton's method iteration to refine the solutions to this equation. The solution near 1.00995 requires the initial guess be very close for that method to work properly. Fortunately, the 1.01 value shown on the graph is sufficient for the purpose.
PLEASE HURRY! i walked north 8 miles, the west 4 miles, and finally south 5 miles, at the end how far was i from where i started
Answer:
5 miles away
Step-by-step explanation:
If you walked north 8 miles, then west 4 miles, then south 5 miles, you have, in total, travelled 4 miles west and [tex]8-5=3[/tex] miles north.
This creates a triangle, in which we can find the the length of the hypotenuse to find how far away you are now.
We can use the Pythagorean theorem since this is a right triangle.
[tex]a^2+b^2=c^2\\3^2+4^2=c^2\\9+16=c^2\\25=c^2\\c=5[/tex]
Hope this helped!
Answer:
5 miles away
Step-by-step explanation:
At a local high school, the student population is growing at 12% a year. If the original population was 242 students, how long will it take the population to reach 300 students? Round to the nearest tenth of a year.
Answer: 2 years
Step-by-step explanation:
The exponential growth function is given by :-
[tex]y=A(1+r)^x[/tex] (i)
, where A = initial value , r = rate of growth and x= time period.
As per given ,
A= 242
r= 12% = 0.12
To find : t when y= 300.
Put all the values in (i)
[tex]300=242(1+0.12)^x\\\\\Rightarrow\ \dfrac{300}{242}=(1.12)^x\\\\\Rightarrow\ 1.23967=(1.12)^x[/tex]
Taking log on both sides , we get
[tex]\log (1.2396) = t \log (1.12)\\\\\Rightarrow\ 0.09328=t(0.049218)\\\\\Rightarrow t=\dfrac{0.09328}{0.049218}=\approx2[/tex]
hence, it will take 2 years.
Hey market sales six cans of food for every seven boxes of food the market sold a total of 26 cans and boxes today how many of each kind did the market sale
Answer:
It sold 14 cans boxes of food and 12 cans of food.
Step-by-step explanation:
The factor for the food cans depend upon every seven food boxes .So, the same no. of sets of food cans will be sold.
Let the no. of sets of food boxes be x.
According to the question,
6x+7x=26
13x=26
x=26/13
x=2
No. of food cans =6x=6×2=12 cans
No. of food boxes=7x=7×2=14 boxes
Please mark brainliest ,if it is truly the best ! Thank you!
During two years in college, a student earned $9,500. The second year she earned $500 more than twice the amount she earned the first year. How much did she earn the first year?
The value of y varies directly with x . Find the value of k when y 33.6 and x = 4.2
Answer:
k=8
Step-by-step explanation:
Since y and x are in direct proportions, the equation is
y= kx, where k is a constant.
when y= 33.6, x=4.2,
33.6= k(4.2)
k= 33.6 ÷4.2
k=8
Answer:
k=8
Step-by-step explanation:
The age of some lecturers are 42,54,50,54,50,42,46,46,48 and 48 calculate the mean age and standard deviation
Answer:
Mean age: 48
Standard deviation: 4
Step-by-step explanation:
a) Mean
The formula for Mean = Sum of terms/ Number of terms
Number of terms
= 42 + 54 + 50 + 54 + 50 + 42 + 46 + 46 + 48+ 48/ 10
= 480/10
= 48
The mean age is 48
b) Standard deviation
The formula for Standard deviation =
√(x - Mean)²/n
Where n = number of terms
Standard deviation =
√[(42 - 48)² + (54 - 48)² + (50 - 48)² +(54 - 48)² + (50 - 48)² +(42 - 48)² + (46 - 48)² + (46 - 48)² + (48 - 48)² + (48 - 48)² / 10]
= √-6² + 6² + 2² + 6² + 2² + -6² + -2² + -2² + 0² + 0²/10
=√36 + 36 + 4 + 36 + 4 + 36 + 4 + 4 + 0 + 0/ 10
=√160/10
= √16
= 4
The standard deviation of the ages is 4
Robert is putting new roofing shingles on his house. Each shingle is 1 2/3 feet long. The north part of the house has a roof line that is 60 feet across. How many shingles can be placed (side by side) on the north part of the house?
Answer: 36 shingles can be placed on the north part of the house.
Step-by-step explanation:
Given: Length of each shingle = [tex]1\dfrac23[/tex] feet = [tex]\dfrac53[/tex] feet.
The north part of the house has a roof line that is 60 feet across.
Then, the number of shingles can be placed on the north part of the house = (Length of roof line in north part) ÷ (Length of each shingle)
[tex]=60\div \dfrac{5}{3}\\\\=60\times\dfrac{3}{5}\\\\=12\times3=36[/tex]
Hence, 36 shingles can be placed on the north part of the house.
