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Doss [256]
2 years ago
8

Determine the slope of the line which contains the following points. (-3,5) and (2,-6)

Mathematics
1 answer:
Feliz [49]2 years ago
7 0
Your slope would be -1. -6-2 divided by 5-(-3) is -8 over 8, simplifying to -1
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Pleaseee helppp yalllll
Sliva [168]

Answer:

285

Step-by-step explanation:

585 - 300 = 285

3 0
2 years ago
Question 3. Y=(1/5)^xSketch the graph of each of the exponential functions and label three points on each graph.
STatiana [176]

Given exponential function:

y\text{ = (}\frac{1}{5})^x

Let us obtain three points including the y-intercept so that we can plot the function y = f(x)

When x =0:

\begin{gathered} y\text{ = (}\frac{1}{5})^0 \\ =\text{ 1} \end{gathered}

when x =1:

\begin{gathered} y\text{ = (}\frac{1}{5})^1 \\ =\text{ }\frac{1}{5} \end{gathered}

when x =2:

\begin{gathered} y\text{ = (}\frac{1}{5})^2 \\ =\text{ }\frac{1}{25} \end{gathered}

We have the points : (0, 1), (1, 1/5), and (2, 1/25)

Using these points, let us provide a sketch of the plot of y =f(x). We have the plot as shown below:

5 0
1 year ago
Oceanside Bike Rental Shop charges a sixteen dollar fixed fee plus seven dollars an
Romashka-Z-Leto [24]

Answer:

8 hours

Step-by-step explanation:

72-16=56

56÷7=8 hours

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
2 years ago
Please help i need answer asap
ycow [4]

Answer:

1: x=-3

2: x=1 is the answer

5 0
2 years ago
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