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goldfiish [28.3K]
3 years ago
8

kendra biked 10kilometers on monday .she biked twice as many kilometers on tuseday how many total meters did she bike

Mathematics
1 answer:
hram777 [196]3 years ago
5 0

Answer:

10km+10km*2 = 30km

30km*(1000m/1km) = 30000m

In the second equation of the problem, the kilometers from the initial total distance biked is cancelled out by the conversion of kilometers to meters, as seen in the 1000m/1km (or 1000 meters for every 1 kilometer).

Read more on Brainly.com - brainly.com/question/454009#readmore

Step-by-step explanation:

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Tema [17]

Answer:

f(8) = \underline{5}

Step-by-step explanation:

We can see it on the given table of values.

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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

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
-12.405 as a mixed number in simplest form please
scZoUnD [109]

\\ \ast\sf\longmapsto -12.405

\\ \ast\sf\longmapsto -\dfrac{12405}{1000}

  • Simplify until possible

\\ \ast\sf\longmapsto -\dfrac{2481}{200}

Now

\\ \ast\sf\longmapsto -12\dfrac{81}{200}

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3 years ago
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Shtirlitz [24]

The equation is P= 2w + 2(w+4.1)

The width is 8.2cm

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One out of every five students in this class has an A. What percent of the students have an A?​
RSB [31]
20% have an A because 1/5 of 100 is 20
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