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KonstantinChe [14]
3 years ago
13

John left his house in Irvine at 8:35 am to drive to a meeting in Los Angeles, 45 miles away. He arrived at the meeting at 9:50.

At 3:30 pm, he left the meeting and drove home. He arrived home at 5:18.
a What was his average speed on the drive from Irvine to Los Angeles?
b What was his average speed on the drive from Los Angeles to Irvine?
c What was the total time he spent driving to and from this meeting?
d John drove a total of 90 miles roundtrip. Find his average speed. (Round to the nearest tenth.)
Mathematics
1 answer:
Vsevolod [243]3 years ago
4 0

Answer:

b is the correct answer. I'm 95% positive

Step-by-step explanation:

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You take $18.80 divide by 2 =$9.40 then $18.80+$9.40=$28.20
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In a geometric sequence, the ____ between consecutive terms is constant.
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The Answer would be: Ratio
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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

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
Part 2. State what additional information is required in order to know that the triangles are congruent by ASA.​
Helga [31]
<h3>Answer: Choice D. \angle SQR \cong \angle XQR</h3>

The red angle markers show those two angles are congruent. That's one "A" of "ASA". The S refers to a congruent pair of sides. We don't have any tickmarks to indicate congruent pairs; however, we do know that QR = QR is a shared side that overlaps (reflexive theorem). So this is the "S" in "ASA".

The thing missing is the angle Q of the top triangle, and also of the bottom triangle as well. If we know those two angles are congruent, then we have enough info to use ASA. More specifically, if we know that \angle SQR \cong \angle XQR, then we can use ASA.

One thing to notice is that the other answer choices involve side lengths and not angles. This implies that if A, B or C were one of the answers, then we would have something like SAS or SSS. But instead we want ASA. So we can immediately rule choices A,B, and C out.

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3 years ago
LAST ONE I SWEAR BUT STILL WILL MARK BRAINLIEST
Svetlanka [38]

Actually I think it may be C

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