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iren [92.7K]
3 years ago
12

Which equation is correct?

Mathematics
2 answers:
Tpy6a [65]3 years ago
8 0

Answer:

B

Step-by-step explanation:

Sin x°= opposite ÷ hypotenuse

Cos x°= adjacent ÷ hypotenuse

Tan x°= opposite ÷ adjacent

Ctg x°= adjacent ÷ opposite

Vlad [161]3 years ago
6 0

Step-by-step explanation:

SOHCAHTOA

sin=opp/hyp

cos=adj/hyp

tan=opp/adj

option B is correct

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Find the median of the following data set 6,2,59,12,11,9,9,54,54,46,2,32,43,11
Aleonysh [2.5K]

By arranging the numbers in ascending order we get,

2, 2, 6, 9, 9, 11, 11, 12, 32, 43, 46, 54, 54, 59

The 2 middle numbers are:

11, 12

To find the median we have to add them and divide them by 2

So, (11+12)÷2

23÷2

11.5

5 0
4 years ago
Read 2 more answers
What is 1/2 multiplied by 3/5
nadya68 [22]
Answer : 1/2 x 3/5 = 3/10.
3 0
4 years ago
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Help please V/5 = 3/7 v=
Serggg [28]

Answer:

v = 15/7, or 2.14

Step-by-step explanation:

Find a common denominator, or multiply both denominators by each other, and well as the numerator.

v/5 = 3/7 can be written as 7v/35 = 15/35

multiplying both sides with 35, we are left with 7v = 15

v = 15/7, or 2.14

8 0
3 years ago
Read 2 more answers
Find the cdf F(x) associated with each of the following probability density functions. Sketch the graphs of f(x) and F(x).
wolverine [178]

Answer:

See steps below

Step-by-step explanation:

a)

\bf f(x)=3(1-x)^2\;(0

\bf F(x)=\int_{0}^{x}f(t)dt=\int_{0}^{x}3(1-t)^2dt=3\int_{0}^{x}(1-t)^2=1-(1-x)^3

The cdf associated with f is

\bf \boxed{F(x)=1-(1-x)^3} for 0<x<1

<h3>See picture 1 </h3>

The median is a point x such that

F(x) = ½

so, the median is

\bf 1-(1-x)^3=1/2\rightarrow (1-x)^3=1/2\rightarrow \boxed{x=1-\sqrt[3]{2}}

The 25th percentile equals the 1st quartile and is a point x such

F(x) = ¼

and the 25th percentile is

\bf 1-(1-x)^3=1/4\rightarrow (1-x)^3=3/4\rightarrow \boxed{x=1-\sqrt[3]{3/4}}

b)

\bf f(x)=\frac{1}{x^2}\;(1

\bf F(x)=\int_{1}^{x}f(t)dt=\int_{1}^{x}\frac{dt}{t^2}=1-\frac{1}{x}

The cdf associated with f is

\bf \boxed{F(x)=1-\frac{1}{x}} for x>1

<h3>See picture 2 </h3>

The median is

\bf 1-\frac{1}{x}=\frac{1}{2}\rightarrow \boxed{x=2}

The 25th percentile is  

\bf 1-\frac{1}{x}=\frac{1}{4}\rightarrow \boxed{x=4/3}

c)

f(x) = 1/3 for 0<x<1 or 2<x<4

\bf F(x)=\int_{0}^{x}\frac{dt}{3}=\frac{x}{3}\;(0

\bf F(x)=\frac{1}{3}+\int_{2}^{x}\frac{dt}{3}=\frac{1}{3}+\frac{x-2}{3}=\frac{x-1}{3}\;(2

The cdf associated with f is

\bf F(x)=\frac{x}{3} for 0<x<1

\bf F(x)=\frac{x-1}{3} for 2<x<4

<h3>See picture 3 </h3>

The median is

\bf \frac{x-1}{3}=1/2\rightarrow x=1+3/2\rightarrow \boxed{x=5/2}

The 25th percentile is  

\bf \frac{x}{3}=1/4\rightarrow \boxed{x=3/4}

4 0
4 years ago
Manny baked 48 cookies for a bake sale. 75% of the cookies were chocolate chip. How many cookies were NOT chocolate chip
klasskru [66]
48 (.75) = 36
48 -36 = 12 cookies or 25%
8 0
3 years ago
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