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polet [3.4K]
3 years ago
11

in a certain normal distribution, 6.3% of the area lies to the left of 36 and 6.3% of the area lies to the right of 42. Find the

mean and the standard deviation ?
Mathematics
1 answer:
KATRIN_1 [288]3 years ago
5 0

As the normal distribution is symmetric the mean is (36 + 42) / 2 = 39

 

p(X < 42) = 1 - 0.063 = 0.937

 

From tables Φ (1.53) = 0.9370

 

From this time:

42 is 1.53 standard deviations above the mean

42 = 39 + 1.53s

1.53s = 3

 

Standard deviation is = 1.96

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ABC underwent a sequence of rigid transformations to give ABC. Which transformations might have taken place ?
SVEN [57.7K]

Either a reflection across the origin. Or a 180° clockwise rotation.

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Determine what type of study is described. Explain. Researchers wanted to determine whether there was an association between hig
victus00 [196]

Answer:

Experimental Study

Step-by-step explanation:

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3 years ago
One urn contains one blue ball (labeled B1) and three red balls (labeled R1, R2, and R3). A second urn contains two red balls (R
marusya05 [52]

Answer:

(a) See attachment for tree diagram

(b) 24 possible outcomes

Step-by-step explanation:

Given

Urn\ 1 = \{B_1, R_1, R_2, R_3\}

Urn\ 2 = \{R_4, R_5, B_2, B_3\}

Solving (a): A possibility tree

If urn 1 is selected, the following selection exists:

B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]

If urn 2 is selected, the following selection exists:

B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]

<em>See attachment for possibility tree</em>

Solving (b): The total number of outcome

<u>For urn 1</u>

There are 4 balls in urn 1

n = \{B_1,R_1,R_2,R_3\}

Each of the balls has 3 subsets. i.e.

B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]

So, the selection is:

Urn\ 1 = 4 * 3

Urn\ 1 = 12

<u>For urn 2</u>

There are 4 balls in urn 2

n = \{B_2,B_3,R_4,R_5\}

Each of the balls has 3 subsets. i.e.

B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]

So, the selection is:

Urn\ 2 = 4 * 3

Urn\ 2 = 12

Total number of outcomes is:

Total = Urn\ 1 + Urn\ 2

Total = 12 + 12

Total = 24

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3 years ago
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meriva

Answer:

The answer is 10 or 10/1

3 0
3 years ago
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