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nignag [31]
2 years ago
14

A math class's mean test score is 88.4. The standard deviation is 4.0. If Kimmie scored 85.9, what is her z-score

Mathematics
1 answer:
kvv77 [185]2 years ago
5 0

The z-score of Kimmie is -0.625

<h3>Calculating z-score</h3>

The formula for calculating the z-score is expressed as;

z = x-η/s

where

η is the mean

s is the standard deviation

x is the Kimmie score

Substitute the given parameter

z = 85.9-88.4/4.0

z = -2.5/4.0

z = -0.625

Hence the z-score of Kimmie is -0.625

Learn more on z-score here: brainly.com/question/25638875

#SPJ1

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Please help me with this!!
Dennis_Churaev [7]

angle QRP=180-(64+47)

=69°

Using sin rule,

p/sinP=q/sinQ.

7.6/sin64=q/sin47

q=6.2cm

Using the formula;

area=1/2p*qsinR

we have;

1/2×7.6×6.2×sin69°=21.995

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7 0
3 years ago
What is the value of the angle ?
Ivan

Answer:

47°

Step-by-step explanation:

The exterior angle of a triangle is equal to the sum of the 2 opposite interior angles

124° is an exterior angle , thus

77 + ? = 124 ( subtract 77 from both sides )

? = 47°

8 0
3 years ago
In △ABC, m∠A=39°, a=11, and b=13. Find c to the nearest tenth.
Talja [164]

For this problem, we are going to use the <em>law of sines</em>, which states:

\dfrac{\sin{A}}{a} = \dfrac{\sin{B}}{b} = \dfrac{\sin{C}}{c}


In this case, we have an angle and two sides, and we are trying to look for the third side. First, we have to find the angle which corresponds with the second side, B. Then, we can find the third side. Using the law of sines, we can find:

\dfrac{\sin{39^{\circ}}}{11} = \dfrac{\sin{B}}{13}


We can use this to solve for B:

13 \cdot \dfrac{\sin{39^{\circ}}}{11} = \sin{B}

B = \sin^{-1}{\Big(13 \cdot \dfrac{\sin{39^{\circ}}}{11}\Big)} \approx 48.1


Now, we can find C:

C = 180^{\circ} - 48.1^{\circ} - 39^{\circ} = 92.9^{\circ}


Using this, we can find c:

\dfrac{\sin{39^{\circ}}}{11} = \dfrac{\sin{92.9^{\circ}}}{c}

c = \dfrac{11\sin{92.9^{\circ}}}{\sin{39^{\circ}}} \approx \boxed{17.5}


c is approximately 17.5.

8 0
2 years ago
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3 years ago
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