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inessss [21]
3 years ago
9

HELP ASAP PLEASE AND THANK YOU

Mathematics
2 answers:
faust18 [17]3 years ago
6 0

Answer:

j+278=296

Step-by-step explanation:

that is the equation because the total is the sum of j and 278 is equal to 296

Rus_ich [418]3 years ago
5 0
Eqn: j+278=276

the question wrote that the TOTAL of j and 278 is the same as 276, which means that j PLUS 278 is EQUAL to 276.

hope this helps :)
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A savings account with $500 now has $120.​
Zepler [3.9K]

Answer:

They spent 380.

$500-$120=$380.

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Need answer ASAP pls it’s important
Sliva [168]

Hey buddy I am here to help!

total outcomes r = 36

multiples of 3 = 3,6

total outcomes of multiple fo 3 = 12

12/36 = 1/3

1/3 is the answer

Hope it helps!

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3 years ago
What’s the answer to this ?
gladu [14]

Answer:

x = 17

Step-by-step explanation:

Since the triangles are similar then corresponding angles are congruent.

∠ I = ∠P ← substitute values and solve for x

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4x + 4 = 72 ( subtract 4 from both sides )

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3 years ago
Pree point sa inyong lahat​
zimovet [89]

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Thanksss

Step-by-step explanation:

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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

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
3 years ago
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