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2 years ago

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A city's mean minimum daily temperature in February is 22 degrees Upper F. Suppose the standard deviation of the minimum temper

ature in February is 5 degrees Upper F and that the distribution of minimum temperatures in February is approximately Normal. What percentage of days in February has minimum temperatures below freezing (32 degrees Upper F )?

1 answer:

Rom4ik [11]2 years ago

3 0

Answer:

97.72%

Step-by-step explanation:

Given that a city's mean minimum daily temperature in February is 22 degrees Upper F, with std deviation 5 degrees.

Let X be the minimum temperature in February

Then X is N(22,5)

To calculate percentage of days in February has minimum temperatures below freezing (32 degrees Upper F )

Probability that X <32 degrees

= $P(Z$

Convert this into percent by multiplying by 100

97.72% of days in February has minimum temperatures below freezing (32 degrees Upper F )

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What is log Subscript 15 Baseline 2 cubed rewritten using the power property?

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Answer:

The equivalent expression

Step-by-step explanation:

Assuming your expression is $\log_{15}2^3$ , then we can use the power property of logarithm , which is $\log_ab^n=n\log_ab$ .

If we let a=15, b=2, and n=3

Then using the power property of logarithm will give us:

$\log_{15}2^3=3\log_{15}2$

Using the power property of logarithm, the equivalent expression is $3\log_{15}2$

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Consider the set of differences, denoted with d, between two dependent sets: 84, 85, 83, 63, 61, 100, 98. Find the sample standa

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Answer:

The sample standard deviation is 15.3.

Step-by-step explanation:

Given data items,

84, 85, 83, 63, 61, 100, 98,

Number of data items, N = 7,

Let x represents the data item,

Mean of the data points,

$\bar{x}=\frac{84+85+83+63+61+100+98}{7}$

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$=\sqrt{\frac{1}{6}\sum_{i=1}^{7} (x_i-82)^2}$

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Let D be the smaller cap cut from a solid ball of radius 8 units by a plane 4 units from the center of the sphere. Express the v

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Answer:

Step-by-step explanation:

The equation of the sphere, centered a the origin is given by $x^2+y^2+z^2 = 64$ . Then, when z=4, we get

$x^2+y^2= 64-16 = 48$ .

This equation corresponds to a circle of radius $4\sqrt[]{3}$ in the x-y plane

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Then, the triple integral that gives us the volume of D in cartesian coordinates is

$\int_{-4\sqrt[]{3}}^{4\sqrt[]{3}}\int_{-\sqrt[]{48-x^2}}^{\sqrt[]{48-x^2}} \int_{4}^{\sqrt[]{64-x^2-y^2}} dz dy dx$ .

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$\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta$ . Recall that the r factor appears because it is the jacobian associated to the change of variable from cartesian coordinates to polar coordinates. This guarantees us that the integral has the same value. (The explanation on how to compute the jacobian is beyond the scope of this answer).

a) For the spherical coordinates, recall that $z = \rho \cos \phi, y = \rho \sin \phi \sin \theta, x = \rho \sin \phi \cos \theta$ . where $\phi$ is the angle of the vector with the z axis, which varies from 0 up to pi. Note that when z=4, that angle is constant over the boundary of the circle we found previously. On that circle. Let us calculate the angle by taking a point on the circle and using the formula of the angle between two vectors. If z=4 and x=0, then y=4 $\sqrt[]{3}$ if we take the positive square root of 48. So, let us calculate the angle between the vector $a=(0,4\sqrt[]{3},4)$ and the vector b =(0,0,1) which corresponds to the unit vector over the z axis. Let us use the following formula

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d ) Let us use the integral in cylindrical coordinates

$\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta=\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} r (\sqrt[]{64-r^2}-4) dr d\theta=\int_{0}^{2\pi} d \theta \cdot \int_{0}^{4\sqrt[]{3}}r (\sqrt[]{64-r^2}-4)dr= 2\pi \cdot (-2\left.r^{2}\right|_0^{4\sqrt[]{3}})\int_{0}^{4\sqrt[]{3}}r \sqrt[]{64-r^2} dr$

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vazorg [7]

Answer:

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Step-by-step explanation:

The height h of the ball is modeled by the following equation

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The problem want you to find the times the ball will be 48 feet above the ground.

It is going to be when:

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We can simplify by 16t. So

$16t(t-2)= 0$

It means that

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