All categories

2 years ago

Hurry help ASAP!!!!!!!!!!! will give brainliest

1 answer:

5 0

1. 3+2+1+2+3+1=12
2. 5+6+6+7+6+6=36
3. 2+4+3+3+2+4=18

Then add all those numbers up and divide them by 18.
12+36+18=66/18=3.6 you can round it up to c.4.

You might be interested in

How do I sovle this Please help

elena-s [515]

Answer:

sup

Step-by-step explanation:

8 0

2 years ago

Plz show or tell me how to show my work for my answer for question 44?

Svetradugi [14.3K]

Which question do you need help with?

8 0

2 years ago

Not sure what to put in they yellow boxes?

Elza [17]

Answer:

Hamburger=31%

Pie=47%

Hot Dog=22%

36=100%

Step-by-step explanation:

What you do is you take the number of how many people have taken the survey and put it over 36.

11/36, 17/36, 8/36=

Hamburger=31%

Pie=47%

Hot Dog=22%

36=100%

8 0

2 years ago

Last Year

Degger [83]

Answer:

Step-by-step explanation:

The computation of the mean is shown below:

= Total sum ÷ number of observations

= (1 + 2 + 2 + 2 + 3 + 1 + 4 + 0 + 5 + 3 + 6 + 2) ÷ 12

= 31 ÷ 12

= 2.58

= 2.6

= 3

5 0

2 years ago

All the fourth-graders in a certain elementary school took a standardized test. A total of 85% of the students were found to be

Aneli [31]

Answer:

There is a 2% probability that the student is proficient in neither reading nor mathematics.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a student is proficient in reading

B is the probability that a student is proficient in mathematics.

C is the probability that a student is proficient in neither reading nor mathematics.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a student is proficient in reading but not mathematics and $A \cap B$ is the probability that a student is proficient in both reading and mathematics.

By the same logic, we have that:

$B = b + (A \cap B)$

Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So

$(A \cup B) + C = 1$