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aliya0001 [1]
3 years ago
14

For a button to fit through its button-hole the hole needs to be the size of the buttons diameter what size buttonhole is needed

for a button with a circumference of 7.38 CM use the value of 3.14 to represent pi
Mathematics
1 answer:
Lena [83]3 years ago
3 0

Answer:the size of the button hole is 2.35 cm

Step-by-step explanation:

For a button to fit through its button-hole, the hole needs to be the size of the buttons diameter.

The formula for determining the circumference of a circle is expressed as

Circumference = πd or 2πr

Where

d represents the diameter of the circle

π is a constant given as 3.14

From the information given, the circumference of the button is 7.38 cm. Therefore, the diameter of the button will be

d = circumference/π = 7.38/3.14

d = 2.35 cm

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The researcher trains a sample of 16 individuals in mindfulness and then collects data on how long they spend eating per day. Th
Mila [183]

Answer:

The calculated t-value is t=0.47.

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>"Americans typically eat quickly and do not spend time enjoying their meal. A researcher plans to test if those trained in mindfulness will take more time to enjoy their meal (that is, they will spend more time eating than the general population). The general population spends an average of 68 minutes per day combined eating their three meals. The researcher trains a sample of 16 individuals in mindfulness and then collects data on how long they spend eating per day. The sample mean for those trained in mindfulness equals 74; the SS = 150. What is the calculated t-value? Round your answer to two decimal places."</em>

The degrees of freedom are (n-1)=(16-1)=15

The standard deviation is:

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The t-value can be calculated as:

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3 years ago
Use Cramer’s Rule to solve system of equation.
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W_y=\left[\begin{array}{ccc}5&4&0\\3&6&2\\7&29&4\end{array}\right]\\\det W_y=5\cdot6\cdot4+3\cdot29\cdot0+4\cdot2\cdot7-7\cdot6\cdot0-29\cdot2\cdot5-3\cdot4\cdot4=-162\\W_z=\left[\begin{array}{ccc}5&2&4\\3&4&6\\7&3&29\end{array}\right]\\\det W_z=5\cdot4\cdot29+3\cdot3\cdot4+6\cdot2\cdot7-7\cdot4\cdot4-3\cdot6\cdot5-3\cdot2\cdot29=324\\\\x=\dfrac{\det W_x}{\det A}=\dfrac{108}{54}=2\\\\y=\dfrac{\det W_y}{\det A}=\dfrac{-62}{54}=-3\\\\z=\dfrac{\det W_z}{\det A}=\dfrac{324}{54}=6

5 0
3 years ago
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