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3 years ago

Please help with any question !! need help

Mathematics

1 answer:

Gre4nikov [31]3 years ago

4 0

Answer:

1/ 4/5

2/ 6/8 = 3/4

3/ 4/5

4/ 4/10 = 2/5

5/ 2/4 = 1/2

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The gardener mows your lawn in $9 and earns $45. Write and solve an equation to find the number of hours of the gardener worked

amid [387]

Answer:

45/9=__

Step-by-step explanation:

Please mark me brainliest

4 0

3 years ago

Help me with this math problem i don’t understand it

pav-90 [236]

Answer:

13meters

Step-by-step explanation:

c^2=a^2+b^2

12^2+5^2=169

find squareroot of 169 which is 13meters

4 0

2 years ago

Please answer with steps shown. Thanks!

maks197457 [2]

Answer: q = 40

Step-by-step explanation:

Given the quadratic formula as

x = [-b +/-√(b^2 -4ac)]/2a

b = -14

a = 1

c = q

Difference d between the two roots.

d = [-b + √(b^2 -4ac)]/2a - [-b -√(b^2 -4ac)]/2a

d = 2√(b^2 -4ac)/2a

d = √(b^2 -4ac)/a

And d = 6

Substituting the values of a,b and c. We have;

6 = √[(-14^2) - (4×1×q)]

Square both sides

6^2 = 196 - 4q

4q = 196 - 36

q = 160/4

q = 40

The equation becomes

x^2 - 14x + 40 = 0

3 0

3 years ago

Hi there can anyone help me with this please

kipiarov [429]

I think it is 64. you add all of them together and hten divide it by 6.

3 0

3 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

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