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3 years ago

Can someone explain why the answer is True. Will make brainliest.

Mathematics

1 answer:

MrRissso [65]3 years ago

8 0

Answer:

True

Step-by-step explanation:

With Sin and Cos, the rule is that the opposites are the same:

SinA=CosB

CosB=SinA

In this problem, it is showing SinA=cosB, so it is the same, it is true.

Hope this helps!

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For brainiest:):):):):):):)

andrew-mc [135]

Answer:

15 = 0.025

16= 3 years

Step-by-step explanation:

15.

I=prt

15= (200)(r)(3)

200(3) = 600r

15 divided by 600r = 40 (0.025)

------------------------------------------------------

16.

I=prt

$t= \frac{I}{Ixp} \\t= \frac{90}{0.05x600} \\t= 3$

t= 3 years

5 0

3 years ago

((Image Included))

Bas_tet [7]

Answer:

=204.13 inches.

Step-by-step explanation:

Using the side x, we can use sine to find the hypotenuse of the triangle with the angle marked 30°.

Sin 30 =x/hypotenuse.

sin 30 = 27/hyp

hyp= 27/sin 30

=54 inches

We can also find the adjacent as follows.

Cos 30 = adjacent/ 54

Adjacent= 54 cos 30

=46.77 inches

Using the angles marked 45 we can find the hypotenuse of the isosceles triangle.

sin 45= x/hypotenuse

sin 45 =27/hypotenuse

hypotenuse = 27/sin 45

=38.18 inches

The hypotenuse of both the triangles making the isosceles triangle are 38.18 inches long.

Perimeter = 54+ 46.77+27+ 38.18+38.18

=204.13 inches.

6 0

2 years ago

Read 2 more answers

Let X be the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. The article "Methodo

Shkiper50 [21]

Answer:

a) $P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.6288$

$P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.1954=0.4335$

b) $P(4\leq X\leq 8)=0.1954+0.1563+0.1042+0.0595+0.0298=0.5452$

c) $P(X \geq 8) = 1-P(X$

d) $P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559$

Step-by-step explanation:

Let X the random variable that represent the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. We know that $X \sim Poisson(\lambda=4)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda=4$ , $Var(X)=\lambda=2$ , $Sd(X)=2$

a. Compute both P(X≤4) and P(X<4).

$P(X\leq 4)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)+P(X=4)$

Using the pmf we can find the individual probabilities like this:

$P(X=0)=\frac{e^{-4} 4^0}{0!}=e^{-4}=0.0183$

$P(X=1)=\frac{e^{-4} 4^1}{1!}=0.0733$

$P(X=2)=\frac{e^{-4} 4^2}{2!}=0.1465$

$P(X=3)=\frac{e^{-4} 4^3}{3!}=0.1954$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.9646$

$P(X< 4)=P(X\leq 3)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)$

$P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.5311=0.7692$

b. Compute P(4≤X≤ 8).

$P(4\leq X\leq 8)=P(X=4)+P(X=5)+ P(X=6)+P(X=7)+P(X=8)$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563$

$P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042$

$P(X=7)=\frac{e^{-4} 4^7}{7!}=0.0595$

$P(X=8)=\frac{e^{-4} 4^8}{8!}=0.0298$

$P(4\leq X\leq 8)=0.1954+0.1563+ 0.1042+0.0595+0.0298=0.5452$

c. Compute P(8≤ X).

$P(X \geq 8) = 1-P(X$

d. What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation?

The mean is 4 and the deviation is 2, so we want this probability

$P(4\leq X \leq 6)=P(X=4)+P(X=5)+P(X=6)$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563$

$P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042$

$P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559$

4 0

3 years ago

Multiply: (−3x − 6)(3x2 − 6x + 3)

zavuch27 [327]

The answer is -9 • (x + 2) • (x - 1)2

7 0

3 years ago

Read 2 more answers

99 POINTS!

laila [671]

Standard deviation = √(n * p * q)

You are given n and p

q = 1 - p

q = 1 -0.2 = 0.8

Standard deviation = √(21 * 0.2 * 0.8)

=√3.36 = 1.83

The answer is A.

6 0

3 years ago

Read 2 more answers