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2 years ago

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The amount of snowfall falling in a certain mountain range is normally distributed with a mean of 105 ' and a standard deviation

of 10 ' what is the probability that the mean annual snowfall during 25 randomly picked years will exceed 107.8 'your answer should be a decimal rounded to the fourth decimal place.

1 answer:

Ksenya-84 [330]2 years ago

7 0

The sample mean is μ=105, and sample standard deviation is σₓ= $\frac{σ}{\sqrt{n}} =\frac{10}{\sqrt{25}} =2$ .

The Z-score is $Z=\frac{ 107.8-105}{2}=1.4$ .

Refer to standard normal distribution table.

The required probability is

$P(X>107.8)=P(Z>1.4)=1-P(Z$

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When the velocity v of an object is very large, the magnitude of the force due to air resistance is proportional to v squared w

Sati [7]

Answer:

Step-by-step explanation:

The model fo the shell is given by the following equation of equilibrium:

$\Sigma F = -b\cdot v^{2} - m\cdot g = m\cdot \frac{dv}{dt}$

This first-order differential equation has separable variables, which are cleared herein:

$\int\limits^t_{0\,s} \, dt = -\frac{m}{b} \int\limits^{0\,\frac{m}{s} }_{600\,\frac{m}{s} } {\frac{1}{ v^{2}+\frac{m}{b}\cdot g } } \, dv$

The solution of this integral is:

$t = -\frac{m}{2b}\cdot \left[\tan^{-1} \left(\frac{v}{\sqrt{\frac{m\cdot g}{b} } }\right) - \tan^{-1} \left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } }\right)\right]$

$\tan^{-1} \left(\frac{v}{\sqrt{\frac{m\cdot g}{b} } } \right)=-\frac{2\cdot b\cdot t}{m} + \tan^{-1}\left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } } \right)$

$\frac{v}{\sqrt{\frac{m\cdot g}{b} } }=\tan \left[-\frac{2\cdot b\cdot t}{m} + \tan^{-1}\left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } } \right)\right]$

$v = \sqrt{\frac{m\cdot g}{b} } \left [\frac{\tan \left(-\frac{2\cdot b \cdot t}{m} \right)+ \left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } } \right)}{1 - \left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } } \right)\cdot \tan \left(-\frac{2\cdot b \cdot t}{m} \right) }\right]$

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2 years ago

A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat

jeyben [28]

Answer:

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.

We are given a function (corrected)

$N(t) = 20(t^2-lnt^2)+ 30$

$N(t) = 20(t^2-2lnt)+ 30$

(a)

First, we take its derivative

$N'(t) = 20(2t-\frac{2}{t})$

Solve N'(t)=0

$20(2t-\frac{2}{t})=0$

Simplifying

$2t^2-2=0$

Solving for t

$t=1\ ,t=-1$

Only t=1 belongs to the valid interval $1\leqslant t\leqslant 15$

Taking the second derivative

$N''(t) = 20(2+\frac{2}{t^2})$

Which is always positive, so t=1 is a minimum

(b)

$N(1)=20(1^2-2ln1)+ 30$

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

$N(15)=20(15^2-2ln15)+ 30$

N(15)=4391.7 is a maximum

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2 years ago

What is log6^e f rewritten using the power property?

marishachu [46]

Given expression : $log_6e^f .$ .

We need to apply power property of logs to rewrite it.

According to log rule of exponents:

$log_b a^n = n log_b(a)$

If we compare given expression with the rule the exponent part is f, base is 6.

Therefore, we need to bring exponent f in front of log.

Therefore, $log_6e^f = flog_6 e$ .

<h3>And correct option is second option $flog_6 e.$ </h3>

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2 years ago

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Why is it not possible to form a right triangle with the lengths 2, 4, and 7? *

puteri [66]

Answer:

No. The sum of 2 and 4 is less than 7.

Step-by-step explanation:

So, first, to create a triangle, the sum of two smallest sides have to be greater than the longest side, which is 2+4 = 6, and 6 is smaller than 7.

So, no, 2, 4, and 7 cannot create a triangle, to be exact a right triangle.

Hope this helps!!

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3 years ago

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tia_tia [17]

The values of BMW after 2 years and 4 years according to the EXPONENTIAL FUNCTION are $33,462 and $20,358.28

Using the parameters given, we define an decreasing exponential function :

$A = A_{0} (1 - r)^{t}$

Where,

$A_{0}$ = initial amount ; r = Rate ; t = time ; A = final amount

Value after 2 years :

t = 2

A = 55000(1 - 0.22)²

A = 55000(0.78)²

A = $33,462

Value after 4 years :

t = 4

A = 55000(1 - 0.22)^4

A = 55000(0.78)^4

A = $20,358.28

Therefore, the value of BMW after 2 years and 4 years $33,462 and $20,358.28 respectively.

Learn more :

brainly.com/question/14355665

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3 years ago