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2 years ago

Mr. canton places 24 desks in 6 equal rows. how many is in each row?? is it 24 times 6

Mathematics

2 answers:

Marizza181 [45]2 years ago

4 0

No you divide 24/6=4 so you have 4 in each row<span />

lisabon 2012 [21]2 years ago

3 0

24/6=4. 4 desks in each row.

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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

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

In which

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

65% were found to be proficient in both reading and mathematics.

This means that $A \cap B = 0.65$

78% were found to be proficient in mathematics

This means that $B = 0.78$

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

$0.78 = b + 0.65$

$b = 0.13$

85% of the students were found to be proficient in reading

This means that $A = 0.85$

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

$0.85 = a + 0.65$

$a = 0.20$

Proficient in at least one:

$(A \cup B) = a + b + (A \cap B) = 0.20 + 0.13 + 0.65 = 0.98$

What is the probability that the student is proficient in neither reading nor mathematics?

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

$C = 1 - (A \cup B) = 1 - 0.98 = 0.02$

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

6 0

2 years ago

In how many ways can you seat 5 women and 5 men in a row if women must seat next to each other,men must seat next to each other,

makvit [3.9K]

Answer: 288ways

Step-by-step explanation:

There are 5men and 5women to be arranged, since the men must seat together, they will be arranged in 5! ways. For the women, since they must also seat together but with siamese twins between them, they can be arranged in 4! ways instead of 5! ways and this is due to presence of the twins among them.

Note that Siamese twins cannot be separated as such both are taken as one making it 4!.

Since the women and men are always sitting together, they can be arranged in 2! ways i.e 2 sexes

The final seating arrangement can be done in 2!×(5!+4!) ways

= 2× (120+24)

= 2×144

= 288ways.

Note that the arrangement of the men and women are added because they can only be arranged differently to ensure different sex are not sitting together.

8 0

2 years ago

Can someone help me with number 7 and 8 pls

Leto [7]

8.romb

Step-by-step explanation:

thats all I know

5 0

2 years ago

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weeeeeb [17]