Jai bought a helmet and a pair of skates.

Mathematics

1 answer:

allochka39001 [22]4 years ago

3 0

Answer:

224

Step-by-step explanation:

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Taylor surveys students in one grade level who own at least one pet. She finds that 50% of the students surveyed own 2 pets, 3 s

Llana [10]

First, we need to work out the total number of students who were being surveyed.

We know that half of the students has two pets. The rest of the students make up the other half. So, we have 3 students + 2 students + 8 students = 13 students that make half of the sample population

That means total number of students being surveyed is 13+13=26 students

Then we work out the probability
P(One pet) = 8/26 = 4/13
P(Two pets) = 1/2
P(Three pets) = 3/26
P( Four pets) = 2/26 = 1/13

The probability distribution is shown in the table below. Let $x$ be the number of pets and $P(X=x)$ is the probability of owning the number of pets

3 0

4 years ago

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Solve using algebraic equation: <br> 5sin2x=3cosx<br><br> (No exponents)

DaniilM [7]

$5\sin2x=3\cos x\iff10\sin x\cos x=3\cos x$

by the double angle identity for sine. Move everything to one side and factor out the cosine term.

$10\sin x\cos x=3\cos x\iff10\sin x\cos x-3\cos x=\cos x(10\sin x-3)=0$

Now the zero product property tells us that there are two cases where this is true,

$\begin{cases}\cos x=0\\10\sin x-3=0\end{cases}$

In the first equation, cosine becomes zero whenever its argument is an odd integer multiple of $\dfrac\pi2$ , so $x=\dfrac{(2n+1)\pi}2$ where $n[/tex ]is any integer.\\Meanwhile,\\[tex]10\sin x-3=0\implies\sin x=\dfrac3{10}$

which occurs twice in the interval $[0,2\pi)$ for $x=\arcsin\dfrac3{10}$ and $x=\pi-\arcsin\dfrac3{10}$ . More generally, if you think of $x$ as a point on the unit circle, this occurs whenever $x$ also completes a full revolution about the origin. This means for any integer $n$ , the general solution in this case would be $x=\arcsin\dfrac3{10}+2n\pi$ and $x=\pi-\arcsin\dfrac3{10}+2n\pi$ .