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poizon [28]
3 years ago
8

Suppose a normal distribution has a mean of 48 and a standard deviation of 2. What is the probability that a data value is betwe

en 46 and 47? Round your answer to the nearest tenth of a percent.
A. 13.0%
B. 14.0%
C. 16.0%
D. 15.0%​
Mathematics
1 answer:
worty [1.4K]3 years ago
8 0

Answer:

c 16.0%

Step-by-step explanation:

You might be interested in
Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'
vodka [1.7K]

Answer:

a) v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9

a(t) = x''(t) = v'(t) =6t-12

b)  0

c) a(t) = x''(t) = v'(t) =6t-12

When the acceleration is 0 we have:

6t-12=0, t =2

And if we replace t=2 in the velocity function we got:

v(t) = 3(2)^2 -12(2) +9=-3

Step-by-step explanation:

For this case we have defined the following function for the position of the particle:

x(t) = t^3 -6t^2 +9t -5 , 0\leq t\leq 10

Part a

From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:

v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9

a(t) = x''(t) = v'(t) =6t-12

Part b

For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

v(t) = 3t^2 -12t +9

We can factorize this function as v(t)= 3 (t^2- 4t +3)=3(t-3)(t-1)

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is 0\leq t\leq 10 we need to analyze the following intervals:

0< t

For this case if we select two values let's say 0.25 and 0.5 we see that

v(0.25) =6.1875, v(0.5)=3.75

And we see that for a=0.5 >0.25=b we have that f(b)>f(a) so then the function is decreasing on this case.  

1

We have a minimum at t=2 since at this value w ehave the vertex of the parabola :

v_x =-\frac{b}{2a}= -\frac{-12}{2*3}= -2

And at t=-2 v(2) = -3 that represent the minimum for this function, we see that if we select two values let's say 1.5 and 1.75

v(1.75) =-2.8125< -2.25= v(1.5) so then the function sis decreasing on the interval 1<t<2

2

We see that the function would be increasing.

3

For this interval we will see that for any two points a,b with a>b we have f(a)>f(b) for example let's say a=3 and b =4

f(a=3) =0 , f(b=4) =9 , f(b)>f(a)

The particle is moving to the right then the velocity is positive so then the answer for this case is: 0

Part c

a(t) = x''(t) = v'(t) =6t-12

When the acceleration is 0 we have:

6t-12=0, t =2

And if we replace t=2 in the velocity function we got:

v(t) = 3(2)^2 -12(2) +9=-3

5 0
3 years ago
Solve using the square root property.X^2=-12
Alla [95]

Answer:

x = ± 2i\sqrt{3}

Step-by-step explanation:

note that \sqrt{-1} = i

x² = - 12 ( take square root of both sides )

x = ± \sqrt{-12} = ± \sqrt{4(3)(-1)} = ± 2i\sqrt{3}

5 0
2 years ago
What is the interquartile range for the data set 12,14,16,16.5,11,18,18,14,20
podryga [215]
Your answer would be 15.6

6 0
3 years ago
PLEASE SOMEONE HELP WITH THESE 2 questions ASAP!! Will give double the points
zalisa [80]
The answer is
2)40
3)y=3\2
5 0
3 years ago
Help me with this question
kaheart [24]
First combine like terms and put the expression in standard form. This expression is now in simplest form and cannot be simplified anymore.
3a + 10ab - 7a^2b - 10ab^2 + 12ab - (-4) + (-a^2b) - 5a
-10ab^2 - 8a^2b -2a + 24ab - (-4)

Hope this helps

5 0
3 years ago
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