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adell [148]
3 years ago
5

*10 ml : 21* pls help me with this ratios question in (simplest form) pls​

Mathematics
1 answer:
larisa86 [58]3 years ago
3 0

Answer:

0,476190476 ml

Step-by-step explanation:

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The average score of all golfers for a particular course has a mean of 61 and a standard deviation of 3.5 . Suppose 49 golfers p
Nadya [2.5K]

Answer:

The probability that the average score of the 49 golfers exceeded 62 is 0.3897

Step-by-step explanation:

The average score of all golfers for a particular course has a mean of 61 and a standard deviation of 3.5

\mu = 61

\sigma = 3.5

We are supposed to find he probability that the average score of the 49 golfers exceeded 62.

Formula : Z=\frac{x-\mu}{\sigma}

Z=\frac{62-61}{3.5}

Z=0.285

Refer the z table for p value

p value = 0.6103

P(x>62)=1-P(x<62)=1-0.6103=0.3897

Hence the probability that the average score of the 49 golfers exceeded 62 is 0.3897

7 0
3 years ago
HELP PLEASE HELP Please
mylen [45]
The answer is b, 5x^2 + 24x + 1
3 0
3 years ago
Nathan has a sculpture in the shape of a pyramid. The height of the sculpture is 3 centimeters less than the side length, x, of
uysha [10]
Let
x-----------> <span>the side length of a pyramid square base
h-----------> t</span>he height of the sculpture <span>in the shape of a pyramid

we know that
h=(x-3)
Volume=162 cm</span>³
Volume=x² *(x-3)/3
then
x² *(x-3)/3=162----------> x³-3x²=486----------> x³-3x²-486=0
x³-3x²-486=0-------- <span>this equation can be used to find the length of the sculpture’s base 

using a graph tool-----------> </span>to find the solution
x=9 cm -------------> see the attached figure
h=(x-3)-----> h=9-3--------> h=6 cm

the answer is
<span>the length of the sculpture’s base is 9 cm
</span>the height of the sculpture is 6 cm

8 0
3 years ago
Identify the domain and range of the function graphed below. <br><br><br><br>​
Vladimir [108]

Answer:

Domain: (-∞, 4]

Range: [0, ∞)

General Formulas and Concepts:

<u>Algebra I</u>

  • Domain is the set of x-values that can be inputted into function f(x)
  • Range is the set of y-values that are outputted by function f(x)

Step-by-step explanation:

According to the graph, we see the line's x-value span from negative infinity to 4. Since 4 is closed dot, it is inclusive in the domain:

(-∞, 4] or x ≤ 4

According to the graph, we see the line's y-value span from 0 to infinity. Since 0 is closed dot, it is inclusive in the range:

[0, ∞) or y ≥ 0

8 0
3 years ago
For the function defined by f(t)=2-t, 0≤t&lt;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
3 years ago
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