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chubhunter [2.5K]
3 years ago
14

I NEED HELP ASAP PLEASE HELP ME!! NO LINKS OR I WILL REPORT

Mathematics
1 answer:
strojnjashka [21]3 years ago
5 0

Answer:

27

Step-by-step explanation:

5(6)-6/2

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There are two thousand girls and one thousand boys at a university.For every girl with glasses there are two boys who were glass
e-lub [12.9K]

Answer:

200

Step-by-step explanation:

2000 girls :1000 boys

2000 +1000 =3000

20% of 3000 = 600

I girl : 2 boys

1+2=3

600÷3 = 200

200 boys wore glasses

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3 years ago
I NEED HELP ASAP PLEASE
11Alexandr11 [23.1K]
The answr is 2 okay that’s the answer
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1 year ago
Find the product:<br> (x+9)(x-9)<br> a 18<br> b. -81<br> C. - 18<br> d. x-64
Lilit [14]

Answer:

x^2 -81

Step-by-step explanation:

(x+9)(x-9)

FOIL

first x*x = x^2

outer -9x

inner 9x

last -9*9 = -81

Add them together

x^2 -9x+9x -81

Combine like terms

x^2 -81

5 0
3 years ago
Read 2 more answers
The distance between flaws on a long cable is exponentially distributed with mean 12 m.
Elden [556K]

Answer:

(a) The probability that the distance between two flaws is greater than 15 m is 0.2865.

(b) The probability that the distance between two flaws is between 8 and 20 m is 0.3246.

(c) The median is 8.322.

(d) The standard deviation is 12.

(e) The 65th percentile of the distances is 12.61 m.

Step-by-step explanation:

The random variable <em>X</em> can be defined as the distance between flaws on a long cable.

The random variable <em>X</em> is exponentially distributed with mean, <em>μ</em> = 12 m.

The parameter of the exponential distribution is:

\lambda=\frac{1}{\mu}=\frac{1}{12}=0.0833

The probability density function of <em>X</em> is:

f_{X}(x)=0.0833e^{-0.0833x};\ x\geq 0

(a)

Compute the  probability that the distance between two flaws is greater than 15 m as follows:

P(X\geq15)=\int\limits^{\infty}_{15}{0.0833e^{-0.0833x}}\, dx\\=0.0833\times \int\limits^{\infty}_{15}{e^{-0.0833x}}\, dx\\=0.0833\times |\frac{e^{-0.0833x}}{-0.0833}|^{\infty}_{15}\\=e^{0.0833\times 15}\\=0.2865

Thus, the probability that the distance between two flaws is greater than 15 m is 0.2865.

(b)

Compute the  probability that the distance between two flaws is between 8 and 20 m as follows:

P(8\leq X\leq20)=\int\limits^{20}_{8}{0.0833e^{-0.0833x}}\, dx\\=0.0833\times \int\limits^{20}_{8}{e^{-0.0833x}}\, dx\\=0.0833\times |\frac{e^{-0.0833x}}{-0.0833}|^{20}_{8}\\=e^{0.0833\times 8}-e^{0.0833\times 20}\\=0.51355-0.1890\\=0.32455\\\approx0.3246

Thus, the probability that the distance between two flaws is between 8 and 20 m is 0.3246.

(c)

The median of an Exponential distribution is given by:

Median=\frac{\ln (2)}{\lambda}

Compute the median as follows:

Median=\frac{\ln (2)}{\lambda}

             =\farc{0.69315}{0.08333}\\=8.322

Thus, the median is 8.322.

(d)

The standard deviation of an Exponential distribution is given by:

\sigma=\sqrt{\frac{1}{\lambda^{2}}}

Compute the standard deviation as follows:

\sigma=\sqrt{\frac{1}{\lambda^{2}}}

   =\sqrt{\frac{1}{0.0833^{2}}}\\=12.0048\\\approx 12

Thus, the standard deviation is 12.

(e)

Let <em>x</em> be 65th percentile of the distances.

Then, P (X < x) = 0.65.

Compute the value of <em>x</em> as follows:

\int\limits^{x}_{0}{0.0833e^{-0.0833x}}\, dx=0.65\\0.0833\times \int\limits^{x}_{0}{e^{-0.0833x}}\, dx=0.65\\0.0833\times |\frac{e^{-0.0833x}}{-0.0833}|^{x}_{0}=0.65\\-e^{-0.0833x}+1=0.65\\-e^{-0.0833x}=-0.35\\-0.0833x=-1.05\\x=12.61

Thus, the 65th percentile of the distances is 12.61 m.

4 0
3 years ago
What is the slope intercept of 7x-1=3y+8
tigry1 [53]

-3

n that's on periodt b

8 0
3 years ago
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