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PtichkaEL [24]
3 years ago
9

shaun baharat and jagan play hockey , shaun has scored 9 more goals that jagan baharat has scored 6 more goals than shaun , all

together they have scored 69 goals , how many did they each score
Mathematics
1 answer:
LekaFEV [45]3 years ago
6 0

Answer:

HTJHJHKMUYI 4643

Step-by-step explanation:

5543544354345455555555555555555555555555555555555555 4

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Each of the 22 members of football team is required to participate in the summer car wash fundraiser the coach estimates that ea
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So since each car wash equals $7, you round 12 to 10 and $7 * 10 = $70.
Plus, 22 to 20.
So, $70 * 20 = $1400
So the answer is 1,400
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1.5 as a mixed number and 1.5 as a improper fraction
sergey [27]
1.5 as a mixed muber is 1 and 1/2, as an improper fraction it could be 3/2
5 0
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Your monthly phone bill is $133.74 and you by 0.25 per mintute your phone cost 9.99 per month. Find the number of mintutes you u
Sever21 [200]
133.74-9.99=123.75
123.75/.25=495
6 0
3 years ago
For the function defined by f(t)=2-t, 0≤t<1, sketch 3 periods and find:
Oksi-84 [34.3K]
The half-range sine series is the expansion for f(t) with the assumption that f(t) is considered to be an odd function over its full range, -1. So for (a), you're essentially finding the full range expansion of the function

f(t)=\begin{cases}2-t&\text{for }0\le t

with period 2 so that f(t)=f(t+2n) for |t| and integers n.

Now, since f(t) is odd, there is no cosine series (you find the cosine series coefficients would vanish), leaving you with

f(t)=\displaystyle\sum_{n\ge1}b_n\sin\frac{n\pi t}L

where

b_n=\displaystyle\frac2L\int_0^Lf(t)\sin\frac{n\pi t}L\,\mathrm dt

In this case, L=1, so

b_n=\displaystyle2\int_0^1(2-t)\sin n\pi t\,\mathrm dt
b_n=\dfrac4{n\pi}-\dfrac{2\cos n\pi}{n\pi}-\dfrac{2\sin n\pi}{n^2\pi^2}
b_n=\dfrac{4-2(-1)^n}{n\pi}

The half-range sine series expansion for f(t) is then

f(t)\sim\displaystyle\sum_{n\ge1}\frac{4-2(-1)^n}{n\pi}\sin n\pi t

which can be further simplified by considering the even/odd cases of n, but there's no need for that here.

The half-range cosine series is computed similarly, this time assuming f(t) is even/symmetric across its full range. In other words, you are finding the full range series expansion for

f(t)=\begin{cases}2-t&\text{for }0\le t

Now the sine series expansion vanishes, leaving you with

f(t)\sim\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi t}L

where

a_n=\displaystyle\frac2L\int_0^Lf(t)\cos\frac{n\pi t}L\,\mathrm dt

for n\ge0. Again, L=1. You should find that

a_0=\displaystyle2\int_0^1(2-t)\,\mathrm dt=3

a_n=\displaystyle2\int_0^1(2-t)\cos n\pi t\,\mathrm dt
a_n=\dfrac2{n^2\pi^2}-\dfrac{2\cos n\pi}{n^2\pi^2}+\dfrac{2\sin n\pi}{n\pi}
a_n=\dfrac{2-2(-1)^n}{n^2\pi^2}

Here, splitting into even/odd cases actually reduces this further. Notice that when n is even, the expression above simplifies to

a_{n=2k}=\dfrac{2-2(-1)^{2k}}{(2k)^2\pi^2}=0

while for odd n, you have

a_{n=2k-1}=\dfrac{2-2(-1)^{2k-1}}{(2k-1)^2\pi^2}=\dfrac4{(2k-1)^2\pi^2}

So the half-range cosine series expansion would be

f(t)\sim\dfrac32+\displaystyle\sum_{n\ge1}a_n\cos n\pi t
f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}a_{2k-1}\cos(2k-1)\pi t
f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}\frac4{(2k-1)^2\pi^2}\cos(2k-1)\pi t

Attached are plots of the first few terms of each series overlaid onto plots of f(t). In the half-range sine series (right), I use n=10 terms, and in the half-range cosine series (left), I use k=2 or n=2(2)-1=3 terms. (It's a bit more difficult to distinguish f(t) from the latter because the cosine series converges so much faster.)

5 0
4 years ago
In a popular online role playing game, players can create detailed designs for their character's "costumes," or appearance. Oliv
sveticcg [70]

Answer:

There is a 54.328% probability that the next person will purchase no more than one costume.

Step-by-step explanation:

Since in a popular online role playing game, players can create detailed designs for their character's "costumes," or appearance, and Olivia sets up a website where players can buy and sell these costumes online, and information about the number of people who visited. the website and the number of costumes purchased in a single day states that 144 visitors purchased no costume, 182 visitors purchased exactly one costume, and 9 visitors purchased more than one costume, to determine, based on these results, the probability that the next person will purchase no more than one costume as a decimal to the nearest hundredth, the following calculation must be performed:

144 + 182 + 9 = 335

335 = 100

182 = X

182 x 100/335 = X

18,200 / 335 = X

54,328 = X

Therefore, there is a 54.328% probability that the next person will purchase no more than one costume.

3 0
3 years ago
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