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Anton [14]
3 years ago
11

SOMEONE PLEASE HELP ASAP! Normal distribution models what type of variable? A.discrete random variable B.random variable C.discr

ete continuous variable D.random continuous variable
Mathematics
1 answer:
Firdavs [7]3 years ago
3 0
<h3>Answer: D. random continuous variable</h3>

A normal distribution has its center at the mean and its spread is determined by the standard deviation. For the standard normal distribution, its mean is 0 and standard deviation is 1. It is in the shape of a bell curve. Most of the values fall in or around 0, forming the peak around this area. As you move away from 0, the values are less frequent which explains the drop-off. The distribution is continuous as you can have any real number (decimal or whole). It's also random because of the nature of statistics.

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Rearrange the equation so u is the independent variable.
vfiekz [6]

Answer: U = 8w-16/-12

Step-by-step explanation:

-12u + 13 = 8w -3

-12u + 13 - 13 = 8w -3 -13

-12u = 8w -16

u = 8w - 16/ -12

4 0
4 years ago
MELISSA WROTE AN INTEGER. THE OPPOSITE OF HER INTEGER IS -67.
Luda [366]

Answer:The integer is 67

Step-by-step explanation:

4 0
3 years ago
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What is the answer to this ??
igomit [66]

Answer:

the first one

Step-by-step explanation:

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3 0
3 years ago
???????? what’s the answer
Len [333]

Answer:

64

Step-by-step explanation:

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8 0
2 years ago
If ABC is an isosceles triangle and DBE is an equilateral triangle find each missing measure
sineoko [7]

Answer:

∠1=∠9=43°,  ∠2=∠7= 17°,  ∠4=∠5=∠6=60°, ∠3=∠8=120°

Step-by-step explanation:

Given ABC is an isosceles triangle and DBE is an equilateral triangle. we have to find each missing measure.

In triangle ABC,

∠1=∠9    (∵isosceles triangle)

⇒ 4x+3=9x-47

⇒ 9x-4x=3+47 ⇒ 5x=50 ⇒ x=10

Hence, ∠9=∠1=(4x+3)°=4(10)+3=43°

Also, given ΔBDE is an equilateral triangle, and all angle of equilateral triangle are equal.

∴ ∠4+∠5+∠6=180°

⇒ ∠4+∠4+∠4=180° ⇒ 3∠4 = 180° ⇒ ∠4 = 60°

∴ ∠4=∠5=∠6=60°

By exterior angle property, ∠3=∠5+∠6=60°+60°=120°

                                             ∠8=∠5+∠4=60°+60°=120°

In ΔABD, ∠1+∠2+∠3=180°

             ⇒ 43°+120°+∠2=180°⇒ ∠2 = 17°

In ΔABD, ∠9+∠8+∠7=180°

             ⇒ 43°+120°+∠7=180°⇒ ∠7= 17°





           

6 0
3 years ago
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