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3 years ago

5% of what number is 63?

1 answer:

6 0

To find the percentage of a number, you take the percent - 5% - and change it to a decimal by moving the percent over 2 times to the left and multiplying it by the number. So...

.05 x 63
= 3.15

Answer: 5% of 63 is 3.15

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Which type of graph is MOST appropriate for making comparisons among data?

worty [1.4K]

Line graph, in physics we would always use line graphs

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4 years ago

Read 2 more answers

Cody bought two burgers and two

8_murik_8 [283]

Answer:

Step-by-step explanation:

Let the cost of one burger be $b while the cost of one small fries be s

2 burgers and 2 small fries cost $14

This means that;

2b + 2s = 14

divide both sides by 2

b + s = 7 •••••••• (i)

Three burgers and four small fries cost $23

Mathematically;

3b + 4s = 23 •••••••• (ii)

From i , we can see that s = 7- b

we can now substitute this into equation ii

3b + 4(7-b) = 23

3b + 28 -4b = 23

4b -3b = 28 -23

b = $5

one burger costs $5

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4 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

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

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.

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3 years ago

What is are indices

White raven [17]

Answer:

The index of a number says how many times to use the number in a multiplication. It is written as a small number to the right and above the base number. In this example: 82 = 8 × 8 = 64. The plural of index is indices. (Other names for index are exponent or power.)

Step-by-step explanation:

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3 years ago

M < TYV=64Â° What is m A.116Â° B.90Â° C.26Â° D.64Â°

solmaris [256]

My answer to this question C

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3 years ago