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3 years ago

5

< < 7 $113103 8 9 10 Evaluate a + b for a = 2 and b = 3.

1 answer:

Lelechka [254]3 years ago

8 0

Answer:

it would be 2+3=5

Step-by-step explanation:

a is 2 and b is 3 so then you add them and will be 5

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Suppose a random variable x is best described by a uniform probability distribution with range 22 to 55. Find the value of a tha

const2013 [10]

Answer:

(a) The value of <em>a</em> is 53.35.

(b) The value of <em>a</em> is 38.17.

(c) The value of <em>a</em> is 26.95.

(d) The value of <em>a</em> is 25.63.

(e) The value of <em>a</em> is 12.06.

Step-by-step explanation:

The probability density function of <em>X</em> is:

$f_{X}(x)=\frac{1}{55-22}=\frac{1}{33}$

Here, 22 < X < 55.

(a)

Compute the value of <em>a</em> as follows:

$P(X\leq a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.95\times 33=[x]^{a}_{22}\\\\31.35=a-22\\\\a=31.35+22\\\\a=53.35$

Thus, the value of <em>a</em> is 53.35.

(b)

Compute the value of <em>a</em> as follows:

$P(X< a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.49\times 33=[x]^{a}_{22}\\\\16.17=a-22\\\\a=16.17+22\\\\a=38.17$

Thus, the value of <em>a</em> is 38.17.

(c)

Compute the value of <em>a</em> as follows:

$P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.85=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.85\times 33=[x]^{55}_{a}\\\\28.05=55-a\\\\a=55-28.05\\\\a=26.95$

Thus, the value of <em>a</em> is 26.95.

(d)

Compute the value of <em>a</em> as follows:

$P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.89=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.89\times 33=[x]^{55}_{a}\\\\29.37=55-a\\\\a=55-29.37\\\\a=25.63$

Thus, the value of <em>a</em> is 25.63.

(e)

Compute the value of <em>a</em> as follows:

$P(1.83\leq X\leq a)=\int\limits^{a}_{1.83} {\frac{1}{33}} \, dx \\\\0.31=\frac{1}{33}\cdot \int\limits^{a}_{1.83} {1} \, dx \\\\0.31\times 33=[x]^{a}_{1.83}\\\\10.23=a-1.83\\\\a=10.23+1.83\\\\a=12.06$

Thus, the value of <em>a</em> is 12.06.

7 0

3 years ago

An arrow is shot vertically upward from a platform 35 ft high at a rate of 166 ft/sec. When will the arrow hit the ground? Ignor

Komok [63]

Answer:

The arrow will hit the ground approximately 10.5 seconds after launch.

Step-by-step explanation:

The arrow experiments a free fall motion, which is a particular form of uniform accelerated motion due to gravity and in which air friction and effects from Earth's rotation can be neglected. Then, the height as a function of time ( $h(t)$ ), measured in feet, is obtained by the following kinematic formula:

$h(t) = h_{o}+v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}$ (1)

Where:

$h_{o}$ - Initial height of the arrow, measured in feet.

$v_{o}$ - Initial velocity of the arrow, measured in feet per second.

$t$ - Time, measured in seconds.

$g$ - Gravitational acceleration, measured in feet per square second.

If we know that $h(t) = 0\,ft$ , $h_{o} = 35\,ft$ , $v_{o} = 166\,\frac{ft}{s}$ and $g = -32.174\,\frac{m}{s^{2}}$ , then the following polynomial is obtained:

$35+166\cdot t -16.087\cdot t^{2}=0$ (2)

Lastly, we find the roots of the given expression by the Quadratic Formula:

$t_{1}\approx 10.526\,s$ and $t_{2} \approx -0.207\,s$

Time is a positive variable and then we conclude that the only solution that is physically reasonable is approximately 10.526 seconds.

The arrow will hit the ground approximately 10.5 seconds after launch.

6 0

3 years ago