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2 years ago

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The score of golfers for a particular course follows a normal distribution that has a mean of 73 and a standard deviation of 3.

Suppose a golfer played the course today. Find the probability that her score is at least 74. 0.4772

1 answer:

Artemon [7]2 years ago

3 0

Answer:

P(X $\geq$ 74) = 0.3707

Step-by-step explanation:

We are given that the score of golfers for a particular course follows a normal distribution that has a mean of 73 and a standard deviation of 3.

Let X = Score of golfers

So, X ~ N( $\mu=73,\sigma^{2}=3^{2}$ )

The z score probability distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean = 73

$\sigma$ = standard deviation = 3

So, the probability that the score of golfer is at least 74 is given by = P(X $\geq$ 74)

P(X $\geq$ 74) = P( $\frac{X-\mu}{\sigma}$ $\geq$ $\frac{74-73}{3}$ ) = P(Z $\geq$ 0.33) = 1 - P(Z < 0.33)

= 1 - 0.62930 = 0.3707

Therefore, the probability that the score of golfer is at least 74 is 0.3707 .

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Without drawing the graph of the function y=1.2x−7, see if the graph passes through points (-10, 5)

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Step-by-step explanation:

Substitute x with -10 and y with 5.

(5) = 1.2(10) - 7

5 = 12 - 7

This is obviously true, so the graph indeed passes through the point (-10, 5).

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2 years ago

The temperature at a point (x, y) on a flat metal plate is given by T(x, y) = 88/(2 + x2 + y2), where T is measured in °C and x,

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Answer:

$D_uT(3,1)=-\frac{44}{9}*\frac{1}{\sqrt{2} } \approx-3.46$

Step-by-step explanation:

To find the rate of change of temperature with respect to distance at the point (3, 1) in the x-direction and the y-direction we need to find the Directional Derivative of T(x,y). The definition of the directional derivative is given by:

$D_uT(x,y)=T_x(x,y)i+T_y(x,y)j$

Where i and j are the rectangular components of a unit vector. In this case, the problem don't give us additional information, so let's asume:

$i=\frac{1}{\sqrt{2} }$

$j=\frac{1}{\sqrt{2} }$

So, we need to find the partial derivative with respect to x and y:

In order to do the things easier let's make the next substitution:

$u=2+x^2+y^2$

and express T(x,y) as:

$T(x,y)=88*u^{-1}$

The partial derivative with respect to x is:

Using the chain rule:

$\frac{\partial u}{\partial x}=2x$

Hence:

$T_x(x,y)=88*(u^{-2})*\frac{\partial u}{\partial x}$

Symplying the expression and replacing the value of u:

$T_x(x,y)=\frac{-176x}{(2+x^2+y^2)^2}$

The partial derivative with respect to y is:

Using the chain rule:

$\frac{\partial u}{\partial y}=2y$

Hence:

$T_y(x,y)=88*(u^{-2})*\frac{\partial u}{\partial y}$

Symplying the expression and replacing the value of u:

$T_y(x,y)=\frac{-176y}{(2+x^2+y^2)^2}$

Therefore:

$D_uT(x,y)=(\frac{1}{\sqrt{2} } )*(\frac{-176x}{(2+x^2+y^2)^2} -\frac{176y}{(2+x^2+y^2)^2})$

Evaluating the point (3,1)

$D_uT(3,1)=(\frac{1}{\sqrt{2} } )*(\frac{-176(3)-176(1)}{(2+3^2+1^2)^2})=(\frac{1}{\sqrt{2} })* (-\frac{704}{144})=(\frac{1}{\sqrt{2} }) ( - \frac{44}{9})\approx -3.46$

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2 years ago

Instructions:Select the correct answer from each drop-down menu. ∆ABC has vertices at A(11, 6), B(5, 6), and C(5, 17). ∆XYZ has

Akimi4 [234]

to compare the triangles, first we will determine the distances of each side

<span>Distance = ((x2-x1)^2+(y2-y1)^2)^0.5
</span>Solving

<span>∆ABC A(11, 6), B(5, 6), and C(5, 17)</span>

<span>AB = 6 units BC = 11 units AC = 12.53 units
</span><span>∆XYZ X(-10, 5), Y(-12, -2), and Z(-4, 15)
</span><span>XY = 7.14 units YZ = 18.79 units XZ = 11.66 units</span>

<span>∆MNO M(-9, -4), N(-3, -4), and O(-3, -15).</span>

<span>MN = 6 units   NO = 11 units MO = 12.53 units
</span><span>∆JKL J(17, -2), K(12, -2), and L(12, 7).
</span><span>JK = 5 units   KL = 9 units JL = 10.30 units
</span><span>∆PQR P(12, 3), Q(12, -2), and R(3, -2)
</span><span>PQ = 5 units   QR = 9 units PR = 10.30 units</span>
Therefore
<span>we have the <span>∆ABC and the </span><span>∆MNO </span><span>
with all three sides equal</span> ---------> are congruent
</span><span>we have the <span>∆JKL </span>and the <span>∆PQR
</span>with all three sides equal ---------> are congruent </span>

let's check

Two plane figures are congruent if and only if one can be obtained from the other by a sequence of rigid motions (that is, by a sequence of reflections, translations, and/or rotations).

1) If ∆MNO ---- by a sequence of reflections and translation --- It can be obtained ------->∆ABC

<span> then </span>∆MNO<span> ≅</span> <span>∆ABC </span>

a) Reflexion (x axis)

The coordinate notation for the Reflexion is (x,y)---- >(x,-y)

<span>∆MNO M(-9, -4), N(-3, -4), and O(-3, -15).</span>

<span>M(-9, -4)-----------------> M1(-9,4)</span>

N(-3, -4)------------------ > N1(-3,4)

O(-3,-15)----------------- > O1(-3,15)

b) Reflexion (y axis)

The coordinate notation for the Reflexion is (x,y)---- >(-x,y)

<span>∆M1N1O1 M1(-9, 4), N1(-3, 4), and O1(-3, 15).</span>

<span>M1(-9, -4)-----------------> M2(9,4)</span>

N1(-3, -4)------------------ > N2(3,4)

O1(-3,-15)----------------- > O2(3,15)

c) Translation

The coordinate notation for the Translation is (x,y)---- >(x+2,y+2)

<span>∆M2N2O2 M2(9,4), N2(3,4), and O2(3, 15).</span>

<span>M2(9, 4)-----------------> M3(11,6)=A</span>

N2(3,4)------------------ > N3(5,6)=B

O2(3,15)----------------- > O3(5,17)=C

<span>∆ABC A(11, 6), B(5, 6), and C(5, 17)</span>

∆MNO reflection------- > ∆M1N1O1 reflection---- > ∆M2N2O2 translation -- --> ∆M3N3O3

The ∆M3N3O3=∆ABC

<span>Therefore ∆MNO ≅ <span>∆ABC - > </span>check list</span>

2) If ∆JKL -- by a sequence of rotation and translation--- It can be obtained ----->∆PQR

<span> then </span>∆JKL ≅ <span>∆PQR </span>

d) Rotation 90 degree anticlockwise

The coordinate notation for the Rotation is (x,y)---- >(-y, x)

<span>∆JKL J(17, -2), K(12, -2), and L(12, 7).</span>

<span>J(17, -2)-----------------> J1(2,17)</span>

K(12, -2)------------------ > K1(2,12)

L(12,7)----------------- > L1(-7,12)

e) translation

The coordinate notation for the translation is (x,y)---- >(x+10,y-14)

<span>∆J1K1L1 J1(2, 17), K1(2, 12), and L1(-7, 12).</span>

<span>J1(2, 17)-----------------> J2(12,3)=P</span>

K1(2, 12)------------------ > K2(12,-2)=Q

L1(-7, 12)----------------- > L2(3,-2)=R

∆PQR P(12, 3), Q(12, -2), and R(3, -2)

∆JKL rotation------- > ∆J1K1L1 translation -- --> ∆J2K2L2=∆PQR

<span>Therefore ∆JKL ≅ <span>∆PQR - > </span><span>check list</span></span>

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Answer:

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Step-by-step explanation:

<u>Proportions </u>

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