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Dafna11 [192]
2 years ago
15

Question 4 options:

Mathematics
1 answer:
andrew11 [14]2 years ago
4 0

I. The value of x is equal to 8.

II. The dimensions of the rectangle are 13 and 5 inches respectively.

<u>Given the following data:</u>

  • Area of a rectangle = 65 square inches.
  • Length of rectangle = x + 5 inches
  • Width of rectangle = x - 3 inches

To find the value of x, and the dimensions of the rectangle:

Mathematically, the area of a rectangle is given by the formula:

A = LW

Substituting the values, we have:

65 = (x+5)(x-3)\\\\65=x^2-3x+5x-15\\\\65=x^2+2x-15-65\\\\x^2 +2x-80=0\\\\x^2 +10x-8x-80=0\\\\x(x+10)-8(x+10)=0\\\\(x-8)(x+10)=0

x = 8 or x = -10

<u>For the</u><u> length:</u>

when x = 8

Length = 8 +5

Length = 13 inches.

<u>For the</u><u> width:</u>

when x = 8

Length = 8 -3

Length = 5 inches.

Find more information: brainly.com/question/11037225

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Answer:

10/7 feet

Step-by-step explanation:

Let e = Ethan

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multiply both sides of the equation by 2

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Substitute for e in equation 2

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multiply both sides of the equation by 3/7

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Substitute for k in equation 1

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difference in length = 40/7 - 30/7 = 10/7

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In a certain Algebra 2 class of 30 students, 19 of them play basketball and 12 of them play baseball. There are 8 students who p
Alenkinab [10]

Answer:

Probability that a student chosen randomly from the class plays basketball or baseball is  \frac{23}{30} or 0.76

Step-by-step explanation:

Given:

Total number of students in the class = 30

Number of students who plays basket ball = 19

Number of students who plays base ball = 12

Number of students who plays base both the games = 8

To find:

Probability that a student chosen randomly from the class plays basketball or baseball=?

Solution:

P(A \cup B)=P(A)+P(B)-P(A \cap B)---------------(1)

where

P(A) = Probability of choosing  a student playing basket ball

P(B) =  Probability of choosing  a student playing base ball

P(A \cap B) =  Probability of choosing  a student playing both the games

<u>Finding  P(A)</u>

P(A) = \frac{\text { Number of students playing basket ball }}{\text{Total number of students}}

P(A) = \frac{19}{30}--------------------------(2)

<u>Finding  P(B)</u>

P(B) = \frac{\text { Number of students playing baseball }}{\text{Total number of students}}

P(B) = \frac{12}{30}---------------------------(3)

<u>Finding P(A \cap B)</u>

P(A) = \frac{\text { Number of students playing both games }}{\text{Total number of students}}

P(A) = \frac{8}{30}-----------------------------(4)

Now substituting (2), (3) , (4) in (1), we get

P(A \cup B)= \frac{19}{30} + \frac{12}{30} -\frac{8}{30}

P(A \cup B)= \frac{31}{30} -\frac{8}{30}

P(A \cup B)= \frac{23}{30}

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