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vesna_86 [32]
2 years ago
13

Help asap! giving brainliest!

Mathematics
1 answer:
saw5 [17]2 years ago
4 0

Answer:

Step-by-step explanation:

2

would you like to know how I got that?

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Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally
storchak [24]

Answer:

a) 0.1587

b) 0.0475

c) 0.7938

Step-by-step explanation:

Let's start defining our random variable.

X : ''Thickness (in mm) of ancient prehistoric Native American pot shards discovered in a Hopi village''

X is modeled as a normal random variable.

X ~ N(μ,σ)

Where μ is the mean and σ is the standard deviation.

To calculate all the probabilities, we are going to normalize the random variable X.

We are going to call to the standard normal distribution ''Z''.

[(X - μ) / σ] ≅ Z

We normalize by subtracting the mean to X and then dividing by standard deviation.

We can find the values of probabilities for Z in a standard normal distribution table.

We are going to call Φ(A) to the normal standard cumulative distribution evaluated in a value ''A''

a)

P(X

P(ZΦ(-1) = 0.1587

b)

P(X>7)=P(\frac{X-4.5}{1.5}>\frac{7-4.5}{1.5})

P(Z>1.666)=1-P(Z\leq 1.666)=

1 - Φ(1.666) = 1 - 0.9525 = 0.0475

c)

P(3

P(-1 Φ(1.666) - Φ(-1) = 0.9525 - 0.1587 = 0.7938

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3 years ago
Dawn simplified a power correctly and came up with a value of 64.
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Hello fellow brainlian, I am pretty sure the answer is A)

Step-by-step explanation:

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Find the exact value of each trigonometric function for the given angle θ.
Kay [80]

Answer:

\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=\dfrac{2}{\sqrt{3}}.

Step-by-step explanation:

The given angle is 240 degrees.

We need to find the exact value of each trigonometric function for the given angle θ.

Since \theta=240, it means θ lies in 3rd quadrant. In 3d quadrant only tan and cot are positive.

\sin (240^\circ)=\sin (180^\circ+60^\circ)=-\sin (60^\circ)=-\dfrac{\sqrt{3}}{2}

\cos (240^\circ)=\cos (180^\circ+60^\circ)=-\cos (60^\circ)=-\dfrac{1}{2}

\tan (240^\circ)=\tan (180^\circ+60^\circ)=\tan (60^\circ)=\sqrt{3}

\cot (240^\circ)=\cot (180^\circ+60^\circ)=\cot (60^\circ)=\dfrac{1}{\sqrt{3}}

\sec (240^\circ)=\sec (180^\circ+60^\circ)=-\sec (60^\circ)=-2

\csc (240^\circ)=\csc (180^\circ+60^\circ)=-\csc (60^\circ)=-\dfrac{2}{\sqrt{3}}

Therefore, \sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=-\dfrac{2}{\sqrt{3}}.

8 0
3 years ago
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