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3 years ago

Find the equation of the line.

Mathematics

1 answer:

Ne4ueva [31]3 years ago

3 0

Answer:

y = 2x + 4

Step-by-step explanation:

y = mx + b

b = y-intercept = 4

m = slope = rise/run = 4/2 = 2

y = 2x + 4

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Can somebody help me please? Thank u! :)

tatuchka [14]

Answer:

1. last choice

2. first choice

Step-by-step explanation:

<h3><u>First</u><u> </u><u>Image</u></h3>

The reason why it is the last choice because when times keep going as distance also keeps increasing as well. But the last choice says that run at the end of track then turn around and run back. That means there must be a decreasing distance and a maximum value as well.

Therefore the only with max point/vertex is the absolute value which matches the last choice.

<h3><u>Second</u><u> </u><u>Image</u></h3>

First choice because the car straight forwards without any change. The car runs with a steady speed but the distance will keep changing without any change of distance.

So the answer is first choice.

6 0

3 years ago

Because gambling is a big business, calculating the odds of a gambler winning or losing in every game is crucial to the financi

IRISSAK [1]

Answer:

a) 0.1165

b) 0.0983

c) 0.000455

d) 0.787

e) 0.767

Step-by-step explanation:

5 bars, 4 lemons, 3 cherries, and a bell

Total = 5+4+3+1 = 13

The probability of getting a bar on a slot, P(Ba) = 5/13 = 0.385

A lemon, P(L) = 4/13 = 0.308

A cherry, P(C) = 3/13 = 0.231

A bell, P(Be) = 1/13 = 0.0769

a) Probability of getting 3 lemons = (4/13) × (4/13) × (4/13) = 256/2197 = 0.1165

b) Probability of getting no fruit symbol

On each slot, there are 4+3 = 7 fruit symbols.

Probability of getting a fruit symbol On a slot = 7/13

Probability of not getting a fruit symbol = 1 - (7/13) = 6/13 = 0.462

Probability of not getting a fruit symbol On the three slots = 0.462 × 0.462 × 0.462 = 0.0983

c) Probability of getting 3 bells, the jackpot = (1/13) × (1/13) × (1/13) = 1/2197 = 0.000455

d) Probability of not getting a bell on the 3 slots

Probability of not getting a bell on one slot = 1 - (1/13) = 12/13 = 0.923

Probability of not getting a bell on the 3 slots = (12/13) × (12/13) × (12/13) = 1728/2197 = 0.787

e) Probability of at least one bar is a sum of probabilities

Note that Probability of getting a bar = 5/13 and probability of not getting a bar = 8/13

1) Probability of getting 1 bar and other stuff on the 2 other slots (this can happen in 3 different orders) = 3 × (5/13)×(8/13)×(8/13) = 960/2197 = 0.437

2) Probability of getting 2 bars and other stuff on the remaining slot (this can also occur in 3 different orders) = 3 × (5/13)×(5/13)×(8/13) = 600/2197 = 0.273

3) Probability of getting 3 bars on the slots machine = (5/13) × (5/13) × (5/13) = 125/2197 = 0.0569

Probability of at least one bar = 0.437 + 0.273 + 0.0569 = 0.7669 = 0.767

5 0

3 years ago

Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x

Serggg [28]

I assume $C$ has counterclockwise orientation when viewed from above.

By Stokes' theorem,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

so we first compute the curl:

$\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k$

$\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k$

Then parameterize $S$ by

$\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k$

where the $z$ -component is obtained from

$1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v$

with $0\le u\le\dfrac\pi2$ and $0\le v\le\dfrac\pi2$ .

Take the normal vector to $S$ to be

$\vec r_v\times\vec r_u=2\cos u\cos v\sin^2v\,\vec\imath+\sin u\sin v\sin(2v)\,\vec\jmath+\cos v\sin v\,\vec k$

Then the line integral is equal in value to the surface integral,

$\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}$

6 0

3 years ago

0.000006 write each number in scientific notation?

andre [41]

6.0 x 10^-5 I think is the answer

4 0

3 years ago

Read 2 more answers

A real estate agent has 19 properties that she shows. She feels that there is a 30% chance of selling any one property during a

netineya [11]

Answer:

$P(X \geq 5)=1-P(X$

We can find the individual probabilities:

$P(X=0)=(19C0)(0.3)^0 (1-0.3)^{19-0}=0.00114$

$P(X=1)=(19C1)(0.3)^1 (1-0.3)^{19-1}=0.0092$

$P(X=2)=(19C2)(0.3)^2 (1-0.3)^{19-2}=0.0358$

$P(X=3)=(19C3)(0.3)^3 (1-0.3)^{19-3}=0.0869$

$P(X=4)=(19C4)(0.3)^4 (1-0.3)^{19-4}=0.1491$

And replacing we got:

$P(X \geq 5) = 1-[0.00114+0.009282+0.0358+0.0869+0.149]= 0.7178$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=19, p=0.3)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

And we want to find this probability:

$P(X \geq 5)$

And we can use the complement rule:

$P(X \geq 5)=1-P(X$

We can find the individual probabilities:

$P(X=0)=(19C0)(0.3)^0 (1-0.3)^{19-0}=0.00114$

$P(X=1)=(19C1)(0.3)^1 (1-0.3)^{19-1}=0.0092$

$P(X=2)=(19C2)(0.3)^2 (1-0.3)^{19-2}=0.0358$

$P(X=3)=(19C3)(0.3)^3 (1-0.3)^{19-3}=0.0869$

$P(X=4)=(19C4)(0.3)^4 (1-0.3)^{19-4}=0.1491$

And replacing we got:

$P(X \geq 5) = 1-[0.00114+0.009282+0.0358+0.0869+0.149]= 0.7178$

4 0

3 years ago