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Finger [1]
3 years ago
5

Tan x + sqrt(3) = - 2 tan x

Mathematics
1 answer:
Svet_ta [14]3 years ago
8 0

9514 1404 393

Answer:

  nπ -π/6 . . . for any integer n

Step-by-step explanation:

  tan(x) +√3 = -2tan(x) . . . . . given

  3tan(x) = -√3 . . . . . . . . . . . add 2tan(x)-√3

  tan(x) = -√3/3 . . . . . . . . . . divide by 3

  x = arctan(-√3/3) = -π/6 . . . . use the inverse tangent function to find x

This is the value in the range (-π/2, π/2). The tangent function repeats with period π, so the set of values of x that will satisfy this equation is ...

  x = n·π -π/6 . . . . for any integer n

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If f(x)=k g(x), what is the value of k
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Step-by-step explanation:

the k in algebra is the y intercept :))

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3 years ago
Determine which of the following logarithms is expanded correctly.
ziro4ka [17]

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A) log(jk2y = 7log i + 2logk

Step-by-step explanation:

7 0
3 years ago
The heights of a random sample of 50 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centime
Minchanka [31]

Answer:

Step-by-step explanation:

Hello!

For me, the first step to any statistics exercise is to determine what is the variable of interest and it's distribution.

In this example the variable is:

X: height of a college student. (cm)

There is no information about the variable distribution. To estimate the population mean you need a variable with at least a normal distribution since the mean is a parameter of it.

The option you have is to apply the Central Limit Theorem.

The central limit theorem states that if you have a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.

As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

The sample size in this exercise is n=50 so we can apply the theorem and approximate the distribution of the sample mean to normal:

X[bar]~~N(μ;σ2/n)

Thanks to this approximation you can use an approximation of the standard normal to calculate the confidence interval:

98% CI

1 - α: 0.98

⇒α: 0.02

α/2: 0.01

Z_{1-\alpha /2}= Z_{1-0.01}= Z_{0.99} =2.334

X[bar] ± Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

174.5 ± 2.334* \frac{6.9}{\sqrt{50} }

[172.22; 176.78]

With a confidence level of 98%, you'd expect that the true average height of college students will be contained in the interval [172.22; 176.78].

I hope it helps!

4 0
3 years ago
Tanya has a garden with a trench around it. The garden is a rectangle with length 2.5m and width 2m. The trench and the garden t
Drupady [299]

Answer:

5.5m^2

the area of just the trench is 5.5 m^2

Step-by-step explanation:

The Area of trench only is the total area(trench and garden) minus the Area of garden

Area of trench only = total area(trench and garden) - Area of garden

Area = length × breadth

Substituting the given dimensions;

Total area = 3.5 × 3 = 10.5 m^2

Area of garden = 2.5×2 = 5 m^2

Area of trench only = 10.5 m^2 - 5 m^2

Area of trench only = 5.5m^2

the area of just the trench is 5.5 m^2

8 0
4 years ago
Read 2 more answers
2. Connor went to the carnival with S22.50. He bought a hot dog and a drink for $3.75, and he wanted to spend
igomit [66]

Answer:

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Step-by-step explanation:

7 0
3 years ago
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