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MatroZZZ [7]
3 years ago
14

Mark is buying different nuts to make a mixed nut platter to serve at a party. He buys 1.2 kilograms of peanuts, 300 grams of al

monds and 40 dekagrams of cashews. What is the total weight in grams of the nuts he purchased? Enter only the number. Do not include unit
Mathematics
1 answer:
Nutka1998 [239]3 years ago
8 0

Answer:

1900

Step-by-step explanation:

1 kg = 1000g

1.2kg = x

<u>1</u><u> </u><u>kgx</u> = <u>1</u><u>2</u><u>0</u><u>0</u><u> </u><u>kg.g</u>

1kg. 1kg

x = 1200 g

1 decagram = 10g

40 decagram = x

<u>1</u><u> </u><u>dg.x</u> = <u>4</u><u>0</u><u>0</u><u> </u><u>dg.g</u>

1 dg. 1 dg

x = 400g

so the total= 1200g +300 + 400g=1900g

without the unit = 1900

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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

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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
What is 3 and 3/4 as a decimal
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It should be 3.75 think of it as a dollar 3/4 of a dollar is .75
5 0
3 years ago
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SCORPION-xisa [38]

Answer:

k

Step-by-step explanation:

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2 years ago
ASAP!! Please help me. I will not accept nonsense answers, but will mark as BRAINLIEST if you answer is correctly with solutions
WINSTONCH [101]

Answer:

\boxed{\mathrm{all \: real \: numbers}}

Step-by-step explanation:

F(x)=2^x

The domain of a function is the set of input values that the function can take.

There are no restrictions on the value of x.

The domain of the function is all real numbers.

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8 0
3 years ago
There are 132 people seated in the school auditorium for an assembly.  There are 6 rows in the auditorium, each with the same nu
Damm [24]
You do 132 ÷ 6 which is 22, so that is the answer. This is because if you the auditorium is filled, then there are a total of 132 seats. If there are 6 rows, then there have to be 22 seats each row to make 132 seats. I hope this helps!
7 0
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