What is the excluded value for y= 4/9x-45

Answers

Answer 1

Answer: y=4/9x-45
9x - 45 = 0
9x = 45
x = 5
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Find two unit vectors orthogonal to both given vectors. i j k, 4i k

Answers

The cross product of two vectors gives a third vector $\mathbf v$ that is orthogonal to the first two.

$\mathbf v=(\vec i+\vec j+\vec k)*(4\,\vec i+\vec k)=\begin{vmatrix}\vec i&\vec j&\vec k\n1&1&1\n4&0&1\end{vmatrix}=\vec i+3\,\vec j-4\,\vec k$

Normalize this vector by dividing it by its norm:

$(\mathbf v)/(\|\mathbf v\|)=(\vec i+3\,\vec j-4\,\vec k)/(√(1^2+3^2+(-4)^2))=\frac1{√(26)}(\vec i+3\,\vec j-4\vec k)$

To get another vector orthogonal to the first two, you can just change the sign and use $-\mathbf v$ .

The circle below is centered at the point 8.4 and has a radius of length 4 what is its equation

Answers

Answer:

Step-by-step explanation:

(x-8)^2 + (y-4)^2= 16

Evaluate. Write your answer as a fraction or whole number.
1/3(1)
=

Answers

The answer is 1/3. Any number/ fraction multiplied by 1 remains the original number. Hope this helps!

Answer:

Step-by-step explanation:

The number of minutes that Samantha waits to catch the bus is uniformly distributed between 0 and 15 minutes. What is the probability that Samantha has to wait less than 4.5 minutes to catch the bus? a. 10%
b. 3%
c. 20%
d. 30%

Answers

Answer:

option d. 30%

Step-by-step explanation:

Data provided in the question:

$a=0$

$b=15$

Now,

The probability density function of the uniform distribution is given as:

$f(x)=(1)/(15-0)$

$f(x)=(1)/(15), 0<x<15$

Therefore,

The required probability will be

$P(X<4.5)=\int_(0)^(4.5) f(x) d x$

$P(X<4.5)=\int_(0)^(4.5) (1)/(15) d x$

$P(X<4.5)=\left((x)/(15)\right)_(0)^(4.5)$

$P(X<4.5)=\left((4.5)/(15)-(0)/(15)\right)$

$P(X<4.5)=0.3$

= 0.3 × 100%

= 30%

Hence,

The answer is option d. 30%

Final answer:

In a uniform distribution, all values have an equal chance of occurring. To find the probability of Samantha waiting less than 4.5 minutes, you divide 4.5 by the total time interval of 15 minutes. The answer is 30%.

Explanation:

The question is about the concept of uniform distribution in probability, which means that each value within the given range has an equal chance of being drawn. In this case, Samantha can wait anywhere between 0 and 15 minutes, so each minute has an equally likely chance of being the waiting time. To find the probability that Samantha has to wait less than 4.5 minutes, we simply divide this time interval by the total time interval. Hence, the calculation is 4.5 / 15 = 0.30 or 30%. So, the answer to the question of the probability that Samantha has to wait less than 4.5 minutes to catch the bus is 30%.

Learn more about Uniform Distribution here:

brainly.com/question/33143435

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Suppose the function P(t) = 437.6(1.031)t is used to model the population of an organism in a specific region after t years. When will the number of organisms be 1000? (Round your answer to three decimal places.)

Answers

Answer: 27.071 years.

Step-by-step explanation:

The given function : $P(t) = 437.6(1.031)^t$ is used to model the population of an organism in a specific region after t years.

To find : t , when P(t)=1000

Substitute P(t)=1000 in the given function , we get

$1000 = 437.6(1.031)^t\n\n\Rightarrow\ (1000)/(437.6)=(1.031)^t\n\n\Rightarrow\ 2.2852=(1.031)^t$

Taking natural log on both sides , we get

$\ln(2.2852)=\ln ((1.031)^t)\n\n\Rightarrow\ \ln(2.2852)=t(\ln (1.031))\n\n\Rightarrow\ t=( \ln(2.2852)/(\ln(1.031))\n\n\Rightarrow\ t=(0.8264535)/(0.0305292)\n\n\Rightarrow\ t=27.070918989\approx27.071$

Hence, The number of organisms will be 1000 after t= 27.071 years.

Ten students begin college at the same time. The probability of graduating in four years is 63%. Which expanded expression shows the first and last terms of the expression used to find the probability that at least six will graduate in four years?