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sveta [45]
3 years ago
7

Which of the following are like radicals? Check all of the boxes that apply

Mathematics
1 answer:
ElenaW [278]3 years ago
8 0
There’s no answer choices to answer
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Determine whether the random variable is discrete or continuous. In each​ case, state the possible values of the random variable
yan [13]

Answer:

a) C. The random variable is discrete. The possible values are xequals​0, ​1, ​2,....

b) B. The random variable is continuous. The possible values are x greater than or equals 0.

Step-by-step explanation:

Discrete variables are countable in a finite amount of time. Continuous variables can be infinitely many. They rather assume values in a range.

<u><em>(a) Is the number of hits to a Web site in a day discrete or​ continuous? </em></u>

C. The random variable is discrete. The possible values are x equals​ 0, ​1, ​2,....

<u><em>(b) Is the weight of a Upper T dash bone steak discrete or​ continuous?</em></u>

B. The random variable is continuous. The possible values are x greater than or equals 0.

8 0
3 years ago
Find each angle measure
tigry1 [53]

Answer:

1=125

2=55

3=55

4=125

5=125

6=55

7=55

8=125

Step-by-step explanation:

5 0
3 years ago
Please Help!!!!
jek_recluse [69]

Answer:

(6, -4)

Step-by-step explanation:

The ice cream shop should be between the school and hospital.

4 0
3 years ago
Read 2 more answers
The Polar Express
Igoryamba

Answer:

for 12-8: 33.33

or

for 8-4: 50

or

for 12-4: 66.67

Step-by-step explanation:

4 0
3 years ago
EasyElectrics supermarket orders light bulbs from suppliers, AA Electronics and AAA Electronics. EasyElectrics purchases 30% of
saw5 [17]

Answer: (a) 0.006

               (b) 0.027

Step-by-step explanation:

Given : P(AA) = 0.3 and P(AAA) = 0.70

Let event that a bulb is defective be denoted by D and not defective be D';

Conditional probabilities given are :

P(D/AA) = 0.02 and P(D/AAA) = 0.03

Thus P(D'/AA) = 1 - 0.02 = 0.98

and P(D'/AAA) = 1 - 0.03 = 0.97

(a) P(bulb from AA and defective) = P ( AA and D)

                                                       = P(AA) x P(D/AA)

                                                       = 0.3 x 0.02 = 0.006

(b) P(Defective) = P(from AA and defective) + P( from AAA and defective)

                         = P(AA) x P(D/AA) + P(AAA) x P(D/AAA)

                         = 0.3(0.02) + 0.70(0.03)

                         = 0.027

3 0
3 years ago
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