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3 years ago

Round the number 2.72603 to 2 decimal places

Mathematics

1 answer:

svetlana [45]3 years ago

3 0

Answer:

2.73

Step-by-step explanation:

2.72(6)03

use 6= 2.73

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A² + 6 from 7<br> plzz answer i gave u half of my points plz

omeli [17]

Answer:

a = 1

Step-by-step explanation:

First, you subtract 6 from both sides and you end up with a² = 1. Since one cannot be squared or square rooted, a = 1. I hope this helps!

7 0

3 years ago

On a biased dice , the probability of getting a 6 is 4/5. The dice is rolled 500 times. How many sixes would you expect to roll?

tatyana61 [14]

Find 4/5 of 500.

4/5*500=400

You will roll 400 sixes.

5 0

3 years ago

PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPp

kherson [118]

1. is just 6

2.3/13

3.325/28=11.607

5.16oz i think

3 0

4 years ago

Find the missing side length of the right triangle. show work.

motikmotik

Answer:

6.32

Step-by-step explanation:

To find the side length on a right angled triangle we follow The Pythagorean Theorem:

${a}^{2} + {b}^{2} = {c}^{2}$

c being the hypothesis (longest side: the hypotenuse is always opposite the right angle)

so..

${6}^{2} + {2}^{2} = {x}^{2} \\ 36 + 4 = {x}^{2} \\ 40 = {x}^{2} \\ \sqrt{40} = \sqrt{ {x}^{2} } \\$

$6.32 = x$

8 0

2 years ago

Read 2 more answers

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

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