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Dima020 [189]
3 years ago
12

What is the probability that a randomly selected Californian age 18 to 64 exercises for at least 150 minutes per week? Illustrat

e by noting all relevant details on the normal curve below, and shading the appropriate region.
HELP!! PLEASE!

Mathematics
1 answer:
Pavlova-9 [17]3 years ago
8 0

Answer:

Step-by-step explanation:

This is a normal distribution question with

Mean = 153

Standard Deviation= 10

Since we know that score

Z_{score}=\frac{x- \mu}{ \sigma }

a)

P(x > 150.0)=?

The z-score at x = 150.0 is,

Z = \frac{150-153}{10} \\\\Z=-0.3

This implies that

P(x > 150.0) = P(z > -0.3)

= 1 - P(z < -0.3) = 1 - 0.3820885778110474

P(x > 150) = 0.6179

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Which basic trigonometric identity would you use to verify that tan x cos x =sin x
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Answer:

\huge\boxed{\tan x=\dfrac{\sin x}{\cos x}}

Step-by-step explanation:

\tan x\cos x=\sin x\\\\\tan x=\dfrac{\sin x}{\cos x}\\\\\text{Therefore}\\\\\tan x\cos x=\dfrac{\sin x}{\cos x}\cdot\cos x\qquad\text{cancel}\ \cos x\\\\\tan x\cos x=\sin x\\\\\text{For}\ x\neq\dfrac{\pi}{2}+k\pi},\ k\in\mathbb{Z}

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A building contractor is to dig a foundation 30 feet​ long, 12 feet​ wide, and 6 feet deep. The contractor pays ​$20 per load fo
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$1800

Step-by-step explanation:

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we need to convert this into yards because a truckload is 8 cubic yards, 2160/3 = 720 cubic yd

One truckload is 8 cubic yards

# of truckloads needed: 720/8 = 90 truckloads

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2 years ago
Twelve dogs and cats receive 77 biscuits.
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8 0
3 years ago
Read 2 more answers
The volume V of an ice cream cone is given by V = 2 3 πR3 + 1 3 πR2h where R is the common radius of the spherical cap and the c
Nuetrik [128]

Answer:

The change in volume is estimated to be 17.20 \rm{in^3}

Step-by-step explanation:

The linearization or linear approximation of a function f(x) is given by:

f(x_0+dx) \approx f(x_0) + df(x)|_{x_0} where df is the total differential of the function evaluated in the given point.

For the given function, the linearization is:

V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh

Taking R_0=1.5 inches and h=3 inches and evaluating the partial derivatives we obtain:

V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh\\V(R, h) = V(R_0, h_0) + (\frac{2 h \pi r}{3}  + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh

substituting the values and taking dx=0.1 and dh=0.3 inches we have:

V(R_0+dR, h_0+dh) =V(R_0, h_0) + (\frac{2 h \pi r}{3}  + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh\\V(1.5+0.1, 3+0.3) =V(1.5, 3) + (\frac{2 \cdot 3 \pi \cdot 1.5}{3}  + 2 \pi 1.5^2)\cdot 0.1 + (\frac{\pi 1.5^2}{3} )\cdot 0.3\\V(1.5+0.1, 3+0.3) = 17.2002\\\boxed{V(1.5+0.1, 3+0.3) \approx 17.20}

Therefore the change in volume is estimated to be 17.20 \rm{in^3}

4 0
3 years ago
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