Answer:
The price of bus after 4 yrs is Rs.1652400
Step-by-step explanation:
Present price of car = Rs.3000000
We are given that the price of bus depreciated the first two yrs by 10%
So, The price after first two years =[tex]3000000(1-0.1)^2=2430000[/tex]
Now the price of bus depreciated by 15%
So, The price after third year = 2430000-0.15(2430000)=2065500
Now the price of bus depreciated by 20%
The price after fourth year =2065500-0.2(2065500)=1652400
Hence the price of bus after 4 yrs is Rs.1652400
X-6 greater then equal to 15 + 8x
Answer:x ≤ 3
Step-by-step explanation:
Answer:
x ≥ -3
Step-by-step explanation:
x - 6 ≥ 15 + 8x
x - 8x ≥ 15 + 6
-7x ≥ 21
x ≥ 21/-7
x ≥ -3
x greater than equal to -3
check:
-3 - 6 ≥ 15 + 8*-3
-9 ≥ 15 - 24
Find the missing side length. Leave your answers radical in simplest form. PLEASE HURRY
Answer:
the answers for x and y are both 12
PLEASE HELP!! (1/5) -50 POINTS-
Answer:
[tex]X=\begin{bmatrix}5&3\\ -3&2\end{bmatrix}[/tex]
Step-by-step explanation:
We are given the following matrix equation, from which we have to isolate X and simplify this value.
[tex]\begin{bmatrix}2&4\\ \:\:\:5&4\end{bmatrix}X\:+\:\begin{bmatrix}-8&-8\\ \:\:\:12&1\end{bmatrix}=\:\begin{bmatrix}-10&6\\ \:\:\:25&24\end{bmatrix}[/tex]
To isolate X, let us first subtract the second matrix, as demonstrated below, from either side. Further simplifying this equation we can multiply either side by the inverse of the matrix being the co - efficient of X, isolating it in the doing.
[tex]\begin{bmatrix}2&4\\ 5&4\end{bmatrix}X=\begin{bmatrix}-10&6\\ 25&24\end{bmatrix}-\begin{bmatrix}-8&-8\\ 12&1\end{bmatrix}[/tex] (Simplify second side of equation)
[tex]\begin{bmatrix}-10&6\\ 25&24\end{bmatrix}-\begin{bmatrix}-8&-8\\ 12&1\end{bmatrix}=\begin{bmatrix}\left(-10\right)-\left(-8\right)&6-\left(-8\right)\\ 25-12&24-1\end{bmatrix}=\begin{bmatrix}-2&14\\ 13&23\end{bmatrix}[/tex] ,
[tex]\begin{bmatrix}2&4\\ 5&4\end{bmatrix}X=\begin{bmatrix}-2&14\\ 13&23\end{bmatrix}[/tex] (Multiply either side by inverse of matrix 1)
[tex]X=\begin{bmatrix}2&4\\ 5&4\end{bmatrix}^{-1}\begin{bmatrix}-2&14\\ 13&23\end{bmatrix}=\begin{bmatrix}5&3\\ -3&2\end{bmatrix}[/tex]
Our solution is hence option c
The X- and y-coordinates of point P are each to be chosen at random from the set of integers 1 through 10.
What is the probability that P will be in quadrant II ?
О
1/10
1/4
1/2
Answer:
Ok, as i understand it:
for a point P = (x, y)
The values of x and y can be randomly chosen from the set {1, 2, ..., 10}
We want to find the probability that the point P lies on the second quadrant:
First, what type of points are located in the second quadrant?
We should have a value negative for x, and positive for y.
But in our set; {1, 2, ..., 10}, we have only positive values.
So x can not be negative, this means that the point can never be on the second quadrant.
So the probability is 0.
the formula s= I dont know how to type that but I really need helppppp
Answer:
[tex] s = \sqrt{30} - 2\sqrt{5} m [/tex]
Step-by-step explanation:
Given:
Formula for side length of cube, [tex] s = \sqrt{\frac{SA}{6} [/tex]
Where, S.A = surface area of a cube, and s = side length.
Required:
Difference in side length between a cube with S.A of 180 m² and a cube with S.A of 120 m²
Solution:
Difference = (side length of cube with 180 m² S.A) - (side length of cube with 120 m² S.A)
[tex]s = (\sqrt{\frac{180}{6}}) - (\sqrt{\frac{120}{6}})[/tex]
[tex] s = (\sqrt{30}) - (\sqrt{20}) [/tex]
[tex] s = \sqrt{30} - \sqrt{4*5} [/tex]
[tex] s = \sqrt{30} - 2\sqrt{5} m [/tex]
Write each expression in a simpler form that is equivalent to the given expression. Let F be a nonzero number. f-4
Answer:
f-4
Step-by-step explanation:
f-4 cannot be simplified
This is the simplest form
Answer:
[tex]\large \boxed{f-4}[/tex]
Step-by-step explanation:
[tex]f-4[/tex]
[tex]\sf f \ is \ a \ nonzero \ number.[/tex]
[tex]\sf The \ expression \ cannot \ be \ simplified \ further.[/tex]
the area of a rectangular park is 7/8 sqaure mile. the length of the park is 3/4 mile. what is the width of the park?
Answer:
7/6
Step-by-step explanation:
since the formula of area is length times width,you have to divide the area by the length to find the width
area=length×width
the width will be
width=area÷length
=7/8÷3/4
7/8×4/3
7/2×1/3
7/6
that's the width you can prove it by multiplying the length times the width to see if you will get 7/8..
I hope this helps
A cola-dispensing machine is set to dispense 11 ounces of cola per cup, with a standard deviation of 1.0 ounce. The manufacturer of the machine would like to set the control limit in such a way that, for samples of 35, 5% of the sample means will be greater than the upper control limit, and 5% of the sample means will be less than the lower control limit.Required:a. At what value should the control limit be set?b. If the population mean shifts to 10.7, what is the probability that the change will be detected?c. If the population mean shifts to 11.7, what is the probability that the change will be detected?
Answer:
a. the control limits should be set at (10.72, 11.28)
b. [tex]\mathbf{P(10.72<x<11.28) = 0.4526}[/tex]
c. [tex]\mathbf{P(10.72<x<11.28) = 0.0065}[/tex]
Step-by-step explanation:
Given that:
population mean μ = 11
standard deviation [tex]\sigma[/tex] = 1.0
sample size n = 35
5% of the sample means will be greater than the upper control limit, and 5% of the sample means will be less than the lower control limit.
Therefore, level of significance ∝ = 0.05+0.05 = 0.10
Critical value for [tex]z_{1-\alpha/2} =z_{1-0.10 /2}[/tex]
[tex]\implies z_{1-0.05} = z_{0.95}[/tex]
Using the EXCEL FORMULA: = NORMSINV (0.95)
z = 1.64
The lower control limit and the upper control limit can be determined by using the respective formulas:
Lower control limit = [tex]\mathtt{\mu - z_{1-\alpha/2} \times \dfrac{\sigma}{\sqrt{n}}}[/tex]
Upper control limit = [tex]\mathtt{\mu + z_{1-\alpha/2} \times \dfrac{\sigma}{\sqrt{n}}}[/tex]
For the lower control limit = [tex]11-1.64 \times \dfrac{1.0}{\sqrt{35}}[/tex]
For the lower control limit = [tex]11-0.27721[/tex]
For the lower control limit = 10.72279
For the lower control limit [tex]\simeq[/tex] 10.72
For the upper control limit = [tex]11+1.64 \times \dfrac{1.0}{\sqrt{35}}[/tex]
For the upper control limit = 11 + 0.27721
For the upper control limit = 11.27721
For the upper control limit [tex]\simeq[/tex] 11.28
Therefore , the control limits should be set at (10.72, 11.28)
b. If the population mean shifts to 10.7, what is the probability that the change will be detected?
i.e
[tex]\mathtt{P(10.72<x<11.28) = P( \dfrac{X- \mu}{\dfrac{\sigma}{\sqrt{n}}}<z < \dfrac{X- \mu}{\dfrac{\sigma}{\sqrt{n}}}})[/tex]
[tex]\mathtt{P(10.72<x<11.28) = P( \dfrac{10.72- 10.7}{\dfrac{1.0}{\sqrt{35}}}<z < \dfrac{11.28- 10.7}{\dfrac{1.0}{\sqrt{35}}}})[/tex]
[tex]\mathtt{P(10.72<x<11.28) = P( \dfrac{0.02}{\dfrac{1.0}{5.916}}<z < \dfrac{0.58}{\dfrac{1.0}{5.916}}})[/tex]
[tex]\mathtt{P(10.72<x<11.28) = P(0.1183<z < 3.4313})[/tex]
[tex]\mathtt{P(10.72<x<11.28) = P(z< 3.4313) - P(z< 0.1183) }[/tex]
Using the EXCEL FORMULA: = NORMSDIST (3.4313) - NORMSDIST (0.118 ); we have:
[tex]\mathbf{P(10.72<x<11.28) = 0.4526}[/tex]
c If the population mean shifts to 11.7, what is the probability that the change will be detected?
i.e
[tex]\mathtt{P(10.72<x<11.28) = P( \dfrac{X- \mu}{\dfrac{\sigma}{\sqrt{n}}}<z < \dfrac{X- \mu}{\dfrac{\sigma}{\sqrt{n}}}})[/tex]
[tex]\mathtt{P(10.72<x<11.28) = P( \dfrac{10.72- 11.7}{\dfrac{1.0}{\sqrt{35}}}<z < \dfrac{11.28- 11.7}{\dfrac{1.0}{\sqrt{35}}}})[/tex]
[tex]\mathtt{P(10.72<x<11.28) = P( \dfrac{-0.98}{\dfrac{1.0}{5.916}}<z < \dfrac{-0.42}{\dfrac{1.0}{5.916}}})[/tex]
[tex]\mathtt{P(10.72<x<11.28) = P(-5.7978<z < -2.48472})[/tex]
[tex]\mathtt{P(10.72<x<11.28) = P(z< -2.48472) - P(z< -5.7978) }[/tex]
Using the EXCEL FORMULA: = NORMSDIST (-2.48472) - NORMSDIST (-5.7978); we have:
[tex]\mathbf{P(10.72<x<11.28) = 0.0065}[/tex]
S varies inversely as G. If S is 8 when G is 1.5, find S when G is 3. a) Write the variation. b) Find S when G is 3.
Step-by-step explanation:
a.
[tex]s \: = \frac{k}{g} [/tex]
[tex]8 = \frac{k}{1.5} [/tex]
[tex]k \: = 1.5 \times 8 = 12[/tex]
[tex]s = \frac{12}{g} [/tex]
b.
[tex]s = \frac{12}{3} [/tex]
s = 4
Decimal divison need answer no pdf
Answer:
1.84
Step-by-step explanation:
If you like my answer than please mark me brainliest thanks
What is an equation of the line that passes through the points (2, -7) and (8, -4)?
Answer:
The answer is
[tex]y = \frac{1}{2} x - 8[/tex]Step-by-step explanation:
To find the equation of the line that passes through two points , first find the slope and then use the formula
y - y1 = m(x - x1)
where m is the slope
(x1 , y1) are any of the points
To find the slope of the line using two points we use the formula
[tex]m = \frac{y2 - y1}{x2 - x1} [/tex]
Slope of the line using points
(2, -7) and (8, -4) is
[tex] \frac{ - 4 + 7}{8 - 2} = \frac{3}{6} = \frac{1}{2} [/tex]Now the equation of the line using point (2 , - 7) and slope 1/2 is
[tex] y + 7 = \frac{1}{2} (x - 2)[/tex][tex]y + 7 = \frac{1}{2} x - 1[/tex][tex]y = \frac{1}{2} x - 1 - 7[/tex]We have the final answer as
[tex]y = \frac{1}{2} x - 8[/tex]Hope this helps you
The hypotenuse of a right triangle is 5 inches long. One of the legs is 1 inch longer than the other. What is the length (in inches) of the longer leg?
Answer: 4 inches
Step-by-step explanation:
1. We gonna find the the length of the right triangle legs using Phitagor theorem.
c²=a²+b² (1) , where c is triangle's hypotenuse
a and b are the triangle's legs.
Let the leg a =x, so leg b=x+1 inches
Now we can write the equation using (1)
25=x²+(x+1)²
25=x²+x²+2*x+1
2*x²+2*x-24=0 ( divide by 2 both sides of the equation)
x²+x-12=0
Find the discriminant D=1+12*4=49
√D=7
x1= (-1+7)/2=3 x2=(-1-7)/2=-4 - x2=-4 not possible so length of the leg can not be negative.
So the shorter leg a=x= 3 inches
The longer leg b=x+1=4 inches
6. Ideal measure of Central tendency and dispersion are:
a. Mean and mean deviation
b. Median and quartile deviation
c. Median and standard deviation
d. Mean and standard deviation
Answer:
median and quartile deviation
A sample of 31 observations is selected from a normal population. The sample mean is 11, and the population standard deviation is 3. Conduct the following test of hypothesis using the 0.05 significance level. H0: μ ≤ 10 H1: μ > 10 Is this a one- or two-tailed test?
Answer:
The test is a two -tailed test
Step-by-step explanation:
From the question we are told that
The sample size is n = 31
The sample mean is [tex]\= x =11[/tex]
The sample standard deviation is [tex]\sigma = 3[/tex]
The null hypothesis is [tex]H_o: \mu \le 10[/tex]
The alternative hypothesis is [tex]H_1 : \mu > 10[/tex]
The level of significance is [tex]\alpha = 0.05[/tex]
The test statistics is mathematically represented as
[tex]t = \frac{ \= x - \mu }{ \frac{\sigma }{\sqrt{n} } }[/tex]
substituting values
[tex]t = \frac{ 11 - 10 }{ \frac{3}{\sqrt{ 31} } }[/tex]
[tex]t = 1.85[/tex]
The p- value is mathematically represented as
[tex]p-value = p( t > 1.856) = 0.0317[/tex]
Looking at the value of [tex]p-value \ and \ \alpha[/tex] we see that [tex]p-value < \alpha[/tex] hence we reject the null hypothesis
Given the that the p value is less than 0.05 it mean the this is a two-tailed test
solve 3/4x+5=-9 please
Answer:
exact form: x=-56/3
mixed number form: -18 2/3
Solve for x by simplifying both sides of the equation, then isolating the variable.
m= -1/2 and the point (3, -6) which is the point -slope form of the equation
Answer:
y+6=-1/2(x-3)
Step-by-step explanation:
Point slope form: y-y1=m(x-x1)
Given that:
m=-1/2 and point (3, -6), you just add these numbers into the equation, and this gives:
y+6=-1/2(x-3)
Hope this helped!
Have a nice day!
Benjamin’s and David’s ages add up to 36 years. The sum of twice their respective ages also add up to 72 years. Find their ages
Answer:
It can be any two numbers that sum up to 36.
eg:18+18,30+6,15+21
Step-by-step explanation:
Given:
Let Benjamin's age be x and David's age y.
x+y=36
2x+2y=72
Solution:
As twice of 36=72, any two numbers that add up to 36 ,will give a sum of 72 after multiplying them with 2.Therefore ,Benjamin's and David's age can be any set of numbers that sum up to 36.
find the HCF of 54 and 81 by prime factorization method. pls send the photo
3║54, 81
3║18, 27
3║9, 9
3║3, 3
3║1, 1
HCF = 27
Must click thanks and mark brainliest
Answer:
Hello,
Step-by-step explanation:
[tex]\begin{array}{c||c}\begin{array}{c|c}54&2\\27&3\\9&3\\3&3\\1\\\end{array}&\begin{array}{c|c}81&3\\27&3\\9&3\\3&3\\1\\\end{array}\end{array}\\\\54=2^1*3^3\\ 81=3^4\\\\\boxed{HCF(54,81)=3^3=27}\\[/tex]
The denominator of a fraction is 30 more than the numerator. The value of the fraction is 3/5. Find the fraction.
Answer:
45
------
75
Step-by-step explanation:
Let x be the value of the numerator and x+30 be the value of the denominator
This is equal to 3/5
x 3
-------- = -------
x+30 5
Using cross products
5x = 3(x+30)
Distribute
5x = 3x+90
Subtract 3x from each side
2x = 90
Divide by 2
x = 45
The fraction is
45
-----
30+45
45
------
75
[tex]\dfrac{x}{x+30}=\dfrac{3}{5}\\\\5x=3(x+30)\\5x=3x+90\\2x=90\\x=45\\\\\dfrac{x}{x+30}=\dfrac{45}{45+30}=\dfrac{45}{75}[/tex]
Evaluate integral _C x ds, where C is
a. the straight line segment x = t, y = t/2, from (0, 0) to (12, 6)
b. the parabolic curve x = t, y = 3t^2, from (0, 0) to (2, 12)
Answer:
a. [tex]\mathbf{36 \sqrt{5}}[/tex]
b. [tex]\mathbf{ \dfrac{1}{108} [ 145 \sqrt{145} - 1]}}[/tex]
Step-by-step explanation:
Evaluate integral _C x ds where C is
a. the straight line segment x = t, y = t/2, from (0, 0) to (12, 6)
i . e
[tex]\int \limits _c \ x \ ds[/tex]
where;
x = t , y = t/2
the derivative of x with respect to t is:
[tex]\dfrac{dx}{dt}= 1[/tex]
the derivative of y with respect to t is:
[tex]\dfrac{dy}{dt}= \dfrac{1}{2}[/tex]
and t varies from 0 to 12.
we all know that:
[tex]ds=\sqrt{ (\dfrac{dx}{dt})^2 + ( \dfrac{dy}{dt} )^2}} \ \ dt[/tex]
∴
[tex]\int \limits _c \ x \ ds = \int \limits ^{12}_{t=0} \ t \ \sqrt{1+(\dfrac{1}{2})^2} \ dt[/tex]
[tex]= \int \limits ^{12}_{0} \ \dfrac{\sqrt{5}}{2}(\dfrac{t^2}{2}) \ dt[/tex]
[tex]= \dfrac{\sqrt{5}}{2} \ \ [\dfrac{t^2}{2}]^{12}_0[/tex]
[tex]= \dfrac{\sqrt{5}}{4}\times 144[/tex]
= [tex]\mathbf{36 \sqrt{5}}[/tex]
b. the parabolic curve x = t, y = 3t^2, from (0, 0) to (2, 12)
Given that:
x = t ; y = 3t²
the derivative of x with respect to t is:
[tex]\dfrac{dx}{dt}= 1[/tex]
the derivative of y with respect to t is:
[tex]\dfrac{dy}{dt} = 6t[/tex]
[tex]ds = \sqrt{1+36 \ t^2} \ dt[/tex]
Hence; the integral _C x ds is:
[tex]\int \limits _c \ x \ ds = \int \limits _0 \ t \ \sqrt{1+36 \ t^2} \ dt[/tex]
Let consider u to be equal to 1 + 36t²
1 + 36t² = u
Then, the differential of t with respect to u is :
76 tdt = du
[tex]tdt = \dfrac{du}{76}[/tex]
The upper limit of the integral is = 1 + 36× 2² = 1 + 36×4= 145
Thus;
[tex]\int \limits _c \ x \ ds = \int \limits _0 \ t \ \sqrt{1+36 \ t^2} \ dt[/tex]
[tex]\mathtt{= \int \limits ^{145}_{0} \sqrt{u} \ \dfrac{1}{72} \ du}[/tex]
[tex]= \dfrac{1}{72} \times \dfrac{2}{3} \begin {pmatrix} u^{3/2} \end {pmatrix} ^{145}_{1}[/tex]
[tex]\mathtt{= \dfrac{2}{216} [ 145 \sqrt{145} - 1]}[/tex]
[tex]\mathbf{= \dfrac{1}{108} [ 145 \sqrt{145} - 1]}}[/tex]
find the straight time pay $7.60 per hour x 40 hours
Answer:
The straight time pay for $ 7.60 per hour and 40 work hours per week is $ 304.
Step-by-step explanation:
Let suppose that worker is suppose to work 8 hours per day, so that he must work 5 days weekly. The straight time is the suppose work time in a week, the pay is obtained after multiplying the hourly rate by the amount of hours per week. That is:
[tex]C = \left(\$\,7,60/hour\right)\cdot (40\,hours)[/tex]
[tex]C = \$\,304[/tex]
The straight time pay for $ 7.60 per hour and 40 work hours per week is $ 304.
How many ways are there to choose 22 croissants with at least one plain croissant, at least two cherry croissants, at least three chocolate croissants, at least one almond croissant, at least two apple croissants, and no more than three broccoli croissants
Answer:
There are 6566 ways to choose 22 croissants with at least one plain croissant, at least two cherry croissants, at least three chocolate croissants, at least one almond croissant, at least two apple croissants, and no more than three broccoli croissants.
Step-by-step explanation:
Given:
There are 5 types of croissants:
plain croissants
cherry croissants
chocolate croissants
almond croissant
apple croissants
broccoli croissants
To find:
to choose 22 croissants with:
at least one plain croissant
at least two cherry croissants
at least three chocolate croissants
at least one almond croissant
at least two apple croissants
no more than three broccoli croissants
Solution:
First we select
At least one plain croissant to lets say we first select 1 plain croissant, 2 cherry croissants, 3 chocolate croissants, 1 almond croissant, 2 apple croissants
So
1 + 2 + 3 + 1 + 2 = 9
Total croissants = 22
So 9 croissants are already selected and 13 remaining croissants are still needed to be selected as 22-9 = 13, without selecting more than three broccoli croissants.
n = 5
r = 13
C(n + r - 1, r)
= C(5 + 13 - 1, 13)
= C(17,13)
[tex]=\frac{17! }{13!(17-13)!}[/tex]
= 355687428096000 / 6227020800 ( 24 )
= 355687428096000 / 149448499200
= 2380
C(17,13) = 2380
C(n + r - 1, r)
= C(5 + 12 - 1, 12)
= C(16,12)
[tex]=\frac{16! }{12!(16-12)!}[/tex]
= 20922789888000 / 479001600 ( 24 )
= 20922789888000 / 11496038400
= 1820
C(16,12) = 1820
C(n + r - 1, r)
= C(5 + 11 - 1, 11)
= C(15,11)
[tex]=\frac{15! }{11!(15-11)!}[/tex]
= 1307674368000 / 39916800 (24)
= 1307674368000 / 958003200
= 1307674368000 / 958003200
= 1365
C(15,11) = 1365
C(n + r - 1, r)
= C(5 + 10 - 1, 10)
= C(14,10)
[tex]=\frac{14! }{10!(14-10)!}[/tex]
= 87178291200 / 3628800 ( 24 )
= 87178291200 / 87091200
= 1001
C(14,10) = 1001
Adding them:
2380 + 1820 + 1365 + 1001 = 6566 ways
What is the equivalent of 27/5 in decimal form?
Answer: 5.4
Step-by-step explanation: 27/5, so 5x5 makes 25 and 2 remaining so 5x0.4=2 so answer is 5+0.4 which equals to 5.4
Vu is three times as old as Wu. In 25 years Wu will be twice as old as Vu. How old is Vu now?
Answer: Vu is 15 years old now.
Step-by-step explanation:
Let present age of WU be x.
Then, the present age of Vu = 3x
Also, After 25 years
Age of Wu = x+25
According to the question:
[tex](x+25)=2(3x)\\\\\Rightarrow\ x+25=6x\\\\\Rightarrow\5x=25\\\\\Rightarrow\ x=5[/tex]
Present age of Vu = 3(5) = 15
Hence, Vu is 15 years old now.
Answer:
j
Step-by-step explanation:j
g Refer to the Number of Motorcycles Narrative} Are these probabilities true or false? ?a. P(X > 1)=0.35 b. P(X ≤ 2)=0.85 c. P(1 ≤ X ≤ 2)=.6 d. P(0 < X < 1)=0 e. P(1 ≤ X < 3)=.6 True False
Answer:
False.
Step-by-step explanation:
Probability distribution is the function which describes the likelihood of possible values assuming a random variable. The variance distribution is the squared value of each the difference by the mean. values of probability are squared and then their sum is taken to calculate variance deviation. The number of motorcycles greater than 1 has probability of 0.35 so the probability of x = 1 must also be 0.35.
a test has 10 multiple-choice questions with 6 choices each, followed by 35 true/false questions. if a student guesses on each equation, how many ways can he answer the questions on the test
Answer:
6¹⁰×2³⁵
Step-by-step explanation:
he has 6 choices for the first multiple choice question.
and for each of those he had again 6 more choices to answer the second question. 6×6 = 36
so, for all 10 multiple choice questions he answer in
6¹⁰ different ways = 60466176 ways
then there are 35 true/false questions, which are Bausch again multiple choice questions but with only 2 options instead of 6.
so we get 2³⁵ different possibilities. a huge number.
and they're possible for each of the 60466176 ways of the multiple choice part.
so, in total we have
6¹⁰×2³⁵ different answer possibilities.
I don’t really get this question
You can put [tex]n[/tex] different elements in order in [tex]n![/tex] different ways.
So, you can visit 12 different cities in [tex]12!=479001600[/tex] different ways.
Answer: 479,001,600
Step-by-step explanation:
There are 12 ways to go to the first place, 11 for the second, ten for the third, and so on. So 12! Means 12x11x10x9x8x7x6x5x4x3x2x1.
Please answer this correctly without making mistakes
Answer:
7/10 mi
Step-by-step explanation:
The total distance is 3 miles = 30/10 miles.
The other distances added gives 7/10+7/10+9/10 = 23/10
Therefore the last hop from Kingwood to Silvergrove is 30/10 - 23/10 = 7/10
Lila is camping with her family. She wants to hike to the lake, go fishing, and hike back before 6:05 P.M. It will take 1 hour and 10 minutes to hike to the lake and 1 hour and 50 minutes to hike back. Lila wants to fish for 3 hours and 10 minutes. What is the latest time Lila can start the hike to the lake?
Lila will need to start the hike at 11: 55 a.m. to be back at exactly 6: 05 p.m. or at 11: 54 a.m. to be before 6: 05 p.m (6: 04 p.m.)
Explanation:
To solve this question, the first step is to calculate how much time does hiking to the lake, go fishing, and go back takes in total. This can be calculated by adding the time of the three activities. This means 1 hour 10 minutes + 3 hours 10 minutes + 1 hour 50 minutes which is equal to 6 hours 10 minutes. The detailed process is shown below.
Add the hours: 1 + 3 + 1 = 5
Add the minutes: 10+50 +10 = 70
Also, because the total of minutes is above 60 (each hour has 60 minutes) it is necessary to subtract 60 minutes and add 1 hour.
5 hours + 1 hour and 70 minutes - 60 minutes = 6 hours and 10 minutes
Now, to solve the question subtract the time of the activities to the time Lila needs to complete all the activities.
6: 05 p.m. - 6 hours and 10 minutes = 11: 55 a.m
You can get this result by substracting first the hours and then the minutes
6: 05 p.m. - 6 hours = 12: 05 p.m.
12: 05 - 10 minutes = 11: 55 a.m.
According to this, Lila will need to start the hike at 11: 55 a.m. to be back at exactly 6: 05 p.m. or at 11: 54 a.m. to be before 6: 05 a.m because if she starts at 11: 54 a.m. she will be back at 6:04, which is a minute before 6:05 p.m.
A group of fitness club members lose a combined total of 28 kilograms in 1 week. There are approximately 2.2 pounds in 1 kilogram. Assuming the weight loss happened at a constant rate, about how many pounds did the group lose each day?
Answer:
8.8 pounds
Step-by-step explanation:
Given the following :
Combined weight loss which occurred within a week = 28 kg
Number of days in a week = 7 days
1 kilogram (kg) = 2.2 pounds
Combined weight loss in pounds that occurs within a week:
Weight loss in kg × 2.2
28kg * 2.2 = 61.6 pounds
Assume weight loss occurred at a constant rate :
Weight lost by the group per day :
(Total weight loss / number of days in a week)
(61.6 pounds / 7)
= 8.8 pounds daily
Answer:
88
Step-by-step explanation:
Found the answer and I am doing the quiz rn lel