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What is an equation for a sine curve with amplitude 2, and period 4pi radians ?

Nezavi [6.7K]

1. a sine curve with amplitude 2, and period 4pi radians

the general equation of the sine curve ⇒⇒ y = a sin (nθ)
where: a is the amplitude and n = 2π/perid
∵ amplitude 2, and period 4pi radians

∴ y = 2 sin (θ/2)

The correct answer is option D. y = 2 sin (θ/2)
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2.The period and amplitude of the function ⇒⇒ y = 5 cos 2θ

comparing with y = a cos nθ

where : a is the amplitude and n = 2π/period
amplitude = 5 , period = 2π/n = 2π/2 = π


The correct answer is option B. Period: pi radians: Amplitude:5

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3. tan (2π/3) = tan 120° = -√3
120° lie in the second quadrant and its reference angle = 180° - 120° = 60°
tan function in the second quadrant is negative
∴ tan 120° = - tan 60 = -√3

The correct answer is C. -sqrt3

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4. Tan 5π/6 = tan 150° = -(√3)/3

150° lies in the second quadrant and its reference angle = 180° - 150° = 30°
tan function in the second quadrant is negative
∴ tan 150° = - tan 30 = -(√3)/3

The correct answer is B.-sqrt3/3

8 0

3 years ago

FIRST ANSWER GETS BRAINLIEST!!!

maksim [4K]

I believe it is A and C.

There are 6 slots on the wheel, so both the numbers in A and E can be divided/multiplied to fit this criterion. For example, 3 times 2 is 6 (the number of gender slots) and 1 times 2 is two, so two out of 6 art teachers will be female.

I hope this helps!!!

8 0

3 years ago

The number of cars running a red light in a day, at a given intersection, possesses a distribution with a mean of 3.6 cars and a

Iteru [2.4K]

Answer:

$X \sim N(3.6,5)$

Where $\mu=3.6$ and $\sigma=5$

Then we have:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

With the following parameters:

$\mu_{\bar X}= 3.6$

$\sigma_{\bar X} = \frac{5}{\sqrt{100}}= 0.5$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the number of cars running a red light of a population, and for this case we know the distribution for X is given by:

$X \sim N(3.6,5)$

Where $\mu=3.6$ and $\sigma=5$

Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

With the following parameters:

$\mu_{\bar X}= 3.6$

$\sigma_{\bar X} = \frac{5}{\sqrt{100}}= 0.5$

5 0

3 years ago

On Saturday, the shop had a sale.

devlian [24]

Answer:

3 + 549=546

546-2=544

Step-by-step explanation:

7 0

3 years ago

M1 Matrix is:

harkovskaia [24]

Answer:

(a)First Row, First Column =1

(b)First Row, second Column =0

(c)Second Row, First Column =0

(d)Second Row, second Column =1

Step-by-step explanation:

Given matrix $M=\left(\begin{array}{ccc}-5&3\\-8&5\end{array}\right)$

The Inverse of a 2X2 matrix

$A=\left(\begin{array}{ccc}a&b\\c&d\end{array}\right)$

can be found using the following:

$A^{-1}=\dfrac{1}{ad-bc} \left(\begin{array}{ccc}d&-b\\-c&a\end{array}\right)$

Therefore:

$M^{-1}=\dfrac{1}{(5*-5)-(3*-8)} \left(\begin{array}{ccc}5&-3\\8&-5\end{array}\right)\\=-1\left(\begin{array}{ccc}5&-3\\8&-5\end{array}\right)\\=\left(\begin{array}{ccc}-5&3\\-8&5\end{array}\right)$

Next, we find the product $M^{-1}M$

$M^{-1}M=\left(\begin{array}{ccc}-5&3\\-8&5\end{array}\right)\left(\begin{array}{ccc}-5&3\\-8&5\end{array}\right)\\=\left(\begin{array}{ccc}-5*-5+3*-8&-5*3+3*5\\-8*-5+5*-8&-8*3+5*5\end{array}\right)\\=\left(\begin{array}{ccc}1&0\\0&1\end{array}\right)$

Therefore:

(a)First Row, First Column =1

(b)First Row, second Column =0

(c)Second Row, First Column =0

(d)Second Row, second Column =1

NOTE: The multiplication of a matrix and its inverse always gives the identity matrix as seen above,

7 0

3 years ago