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3 years ago

When you type your brainly password it sensors it out ********** I just tried, I never knew

Mathematics

1 answer:

larisa [96]3 years ago

3 0

Answer:

lol i remember answering this question

Step-by-step explanation:

<h2>MARK ME BRAINLIEST PLZZZZZZZZZZZZZZZZZZZZZZ</h2>

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N a region, there is a 0.70.7 probability chance that a randomly selected person of the population has brown eyes. assume 1414

taurus [48]

'll use the binomial approach. We need to calculate the probabilities that 9, 10 or 11
people have brown eyes. The probability that any one person has brown eyes is 0.8, 
so the probability that they don't is 1 - 0.8 = 0.2. So the appropriate binomial terms are 

(11 C 9)(0.8)^9*(0.2)^2 + (11 C 10)(0.8)^10*(0.2)^1 + (11 C 11)(0.8)^11*(0.2)^0 = 

0.2953 + 0.2362 + 0.0859 = 0.6174, or about 61.7 %. Since this is over 50%, it 

is more likely than not that 9 of 11 randomly chosen people have brown eyes, at 
least in this region. 

Note that (n C r) = n!/((n-r)!*r!). So (11 C 9) = 55, (11 C 10) = 11 and (11 C 0) = 1.

8 0

3 years ago

The principal has a budget of $225 and expects at least 16 people to attend.

ale4655 [162]

Answer:

$48

Step-by-step explanation:

16 x 3 = 48

7 0

3 years ago

Read 2 more answers

A landscape plan includes a rectangular flowers bed that is surrounded the wood in planks. In the drawing, the rectangular flowe

forsale [732]

Answer:

The perimeter of the drawing: 18 inches

The perimeter of the actual flower bed: 540 inches

If they were multiplied by 24 instead the actual perimeter of the flower bed would be 432 inches. The effect is that it decreases by 108 inches.

Step-by-step explanation:

Taking the 4 inches and 5 inches and multiplying them by 30 gets 120 by 150. If its a rectangle we will add together all 4 sides to get the perimeter of the actual one.

5 0

3 years ago

In a randomly selected sample of 100 students at a University, 81 of them had access to a computer at home. Give the value of th

astra-53 [7]

Answer:

The value of the standard error for the point estimate is of 0.0392.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

In a randomly selected sample of 100 students at a University, 81 of them had access to a computer at home.

This means that $n = 100, p = \frac{81}{100} = 0.81$

Give the value of the standard error for the point estimate.

This is s. So

$s = \sqrt{\frac{0.81*0.19}{100}} = 0.0392$

The value of the standard error for the point estimate is of 0.0392.

7 0

3 years ago

(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$