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3 years ago

15

The table shows the total number of trees cut down by a logging company over a two-week period. What was the average rate of cha

nge between Day 9 and Day 14?

1 answer:

irina [24]3 years ago

7 0

Average rate of change = slope so lets get our y's and x's and plug them into y2-y1 / x2-x1. (14-9)/(308-195) and that equals 5/113

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A game contains 3 bags of green chips 2 bags of red chips and 5 bags og yellow chips. there ard 25 green chips in a bag and 30 r

nika2105 [10]

1 game has 3 bags of green chips ; 2 bags of red chips ; 5 bags of yellow chips
1 bag: 25 green chips
1 bag: 30 red chips

total number of chips:
green: 3 x 25 = 75
red: 2 x 30 = 60
yellow: 75 + 60 = 135 answer to part a.

75 + 60 + 135 = 270 total number of chips.

270 / 54 = 5 players. answer to part b.

Each of the 5 players will have 15 green chips, 12 red chips, and 27 yellow chips, for a total of 54 chips.

8 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

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

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

If B has the points, (-2,0) and (3,4), use the point slope formula to calculate the equation of the line.

pogonyaev

5/4 would be ur answer point slip formula is y^2 - y^1 over x^2 - x^1

5 0

3 years ago

Find the sum.172 + (–167) + (–10) + (–144)

Ne4ueva [31]

The answer is negative 149

3 0

3 years ago

Read 2 more answers

The product of two consecutive negative even integers is 24. What is the smaller of the two numbers?

Karo-lina-s [1.5K]

Answer:

The smaller one is -6, the bigger one is -4.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers