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Zina [86]
3 years ago
12

Solve the equation. 10x+2=32 plz help I'm so ***** can someone explain this to me I'm slow lol

Mathematics
1 answer:
Svetllana [295]3 years ago
8 0

Answer:

x = 3

Step-by-step explanation:

10x + 2 = 32 ( subtract 2 from both sides )

10x = 30 ( divide both sides by 10 )

x = \frac{30}{10} = 3

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According to an​ almanac, 60​% of adult smokers started smoking before turning 18 years old. ​(a) compute the mean and standard
larisa [96]
The given problem describes a binomial distribution with p = 60% = 0.6. Given that there are 400 trials, i.e. n = 400.

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\mu=np=400\times0.6=240

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\sigma=\sqrt{npq} \\  \\ =\sqrt{np(1-p)} \\  \\ =\sqrt{400(0.6)(0.4)} \\  \\ =\sqrt{96}=9.80


b.)  The mean means that in an experiment of 400 adult smokers, we expect on the average to get about 240 smokers who started smoking before turning 18 years.


c.) It would be unusual to observe <span>340 smokers who started smoking before turning 18 years old in a random sample of 400 adult​ smokers because 340 is far greater than the mean of the distribution.
340 is greater than 3 standard deviations from the mean of the distribution.</span>
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3 years ago
What is the remainder when x+ - 2x2 – 3x – 7 is divided by x + 2?
guapka [62]

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x+1.x-7

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2x-7

4 0
3 years ago
Given: WZ perpendicular to WY, WZ bisects VY, and m angle V = 40.
vaieri [72.5K]
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Read 2 more answers
Suppose that the Celsius temperature at the point (x, y) in the xy-plane is T(x, y) = x sin 2y and that distance in the xy-plane
liraira [26]

Missing information:

How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point

P = (\frac{1}{2}, \frac{\sqrt 3}{2})

Answer:

Rate = 0.935042^\circ /cm

Step-by-step explanation:

Given

P = (\frac{1}{2}, \frac{\sqrt 3}{2})

T(x,y) =x\sin2y

r = 1m

v = 2m/s

Express the given point P as a unit tangent vector:

P = (\frac{1}{2}, \frac{\sqrt 3}{2})

u = \frac{\sqrt 3}{2}i - \frac{1}{2}j

Next, find the gradient of P and T using: \triangle T = \nabla T * u

Where

\nabla T|_{(\frac{1}{2}, \frac{\sqrt 3}{2})}  = (sin \sqrt 3)i + (cos \sqrt 3)j

So: the gradient becomes:

\triangle T = \nabla T * u

\triangle T = [(sin \sqrt 3)i + (cos \sqrt 3)j] *  [\frac{\sqrt 3}{2}i - \frac{1}{2}j]

By vector multiplication, we have:

\triangle T = (sin \sqrt 3)*  \frac{\sqrt 3}{2} - (cos \sqrt 3)  \frac{1}{2}

\triangle T = 0.9870 * 0.8660 - (-0.1606 * 0.5)

\triangle T = 0.9870 * 0.8660 +0.1606 * 0.5

\triangle T = 0.935042

Hence, the rate is:

Rate = \triangle T = 0.935042^\circ /cm

3 0
2 years ago
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