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3 years ago

6

At 6:00 AM the temperature was 2 degrees Celsius. At 11:00 AMthe temperature was -11 degrees Celsius. By how many degrees Celsiu

s did the temperature change?

1 answer:

bonufazy [111]3 years ago

6 0

Answer:

-9

Step-by-step explanation:

Its dropping down from 2 to 0 so that is negative then the -11 so it would be -9.

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Suppose that you randomly draw one card from a standard deck of 52 cards. After writing down which card was drawn, you replace t

34kurt

Answer:

The probability is 0.0775

The expected value is 4.75 clubs

The standard deviation is 1.8875 clubs

Step-by-step explanation:

The variable X follows a binomial distribution, because we have n identical and independent events (19 cards) with a probability p of success and 1-p of fail (there is a probability of 1/4 to be club and 3/4 to be diamond, heart or spade). Then, the probability that x of the n cards are club is:

$P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}\\P(x)=\frac{19!}{x!(19-x)!}*0.25^{x}*(0.75)^{19-x}$

So, the probability P of drawing at least 8 clubs is:

P = P(8) + P(9) + P(10) + ... + P(18) + P(19)

Replacing, the values of x, from 8 to 19, on the equation above, we get:

P = 0.0775

Additionally, the expected value E(x) and standard deviationS(x) for this distribution is given by:

E(x)=np = 19(0.25) = 4.75

$S(x)=\sqrt{np(1-p)} =\sqrt{19(0.25)(0.75)} =1.8875$

8 0

3 years ago

How would I do this problem?

kenny6666 [7]

If it's a geometric sequence then:

$a_1=27;\ a_2=27\\\\r=\dfrac{a_2}{a_1}\to r=\dfrac{27}{36}=\dfrac{3}{4}=0.75\\\\a_{n+1}=a_nr\\\\a_3=27\cdot0.75=20.25\ CORRECT$

We calculate the fourth and fifth term of the sequence:

$a_4=a_3r\to a_4=20.25\cdot0.75=15.1875\\\\a_5=a_4r\to a_5=15.1875\cdot0.75=11.390625$

Answer:

In year 4 15.1875 animals.

In year 5 11.390625 animals.

7 0

3 years ago

Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)

babunello [35]

$\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k$

We want to find $f(x,y,z)$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=x^2+y$

$\dfrac{\partial f}{\partial y}=y^2+x$

$\dfrac{\partial f}{\partial z}=ze^z$

Integrating both sides of the latter equation with respect to $z$ tells us

$f(x,y,z)=e^z(z-1)+g(x,y)$

and differentiating with respect to $x$ gives

$x^2+y=\dfrac{\partial g}{\partial x}$

Integrating both sides with respect to $x$ gives

$g(x,y)=\dfrac{x^3}3+xy+h(y)$

Then

$f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)$

and differentiating both sides with respect to $y$ gives

$y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C$

So the scalar potential function is

$\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}$

By the fundamental theorem of calculus, the work done by $\vec F$ along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it $L$ ) in part (a) is

$\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}$

and $\vec F$ does the same amount of work over both of the other paths.

In part (b), I don't know what is meant by "df/dt for F"...

In part (c), you're asked to find the work over the 2 parts (call them $L_1$ and $L_2$ ) of the given path. Using the fundamental theorem makes this trivial:

$\displaystyle\int_{L_1}\vec F\cdot\mathrm d\vec r=f(0,0,0)-f(4,0,0)=-\frac{64}3$

$\displaystyle\int_{L_2}\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(0,0,0)=\frac{67}3+3e^4$

8 0

3 years ago

Number three please !!!

kati45 [8]

I would say the answer is 5, please let me know if it’s right!

4 0

3 years ago

In the equation f(x) = 3(x – h) 2+ k, what does (h,k) tell?

oee [108]

Answer:

(h,k) shows the coordintes of the turning point.

7 0

3 years ago

Read 2 more answers