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Nikolay [14]
3 years ago
14

A survey was conducted with high school students in each grade to see how many prefer math or science. Some of the data are show

n below. A 6-column table with 3 rows. The first column has no label with entries math, science, total. The second column is labeled 9 with entries blank, 40, 63. The third column is labeled 10 with entries 18, blank, 26. The fourth column is labeled 11 with entries blank, 15, 29. The fifth column is labeled 12 with entries blank, 32, 67. The sixth column is labeled total with entries 90, 95, 185. Which statement is true about the joint frequencies in this table?
Mathematics
2 answers:
tatyana61 [14]3 years ago
5 0

Answer:

Fourteen 11th graders prefer math and eight 10th graders prefer science.

lina2011 [118]3 years ago
3 0

Answer:

B.Fourteen 11th graders prefer math and eight 10th graders prefer science

Step-by-step explanation:

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Part A: Solve the following system of authentically showing your work: 3x + 2y = 4 4x + 3y = 7.
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HW 7.4/7.5 help mhhhh
Hatshy [7]

Answer:

1. LI=18

2. x=10

3. Yes (Y)

4. Yes (Y)

5. x=15

6. x=18

7. x=8

8. x=6

   y=6.5

Step-by-step explanation:

1. LI/JL=KH/JK

Replacing the given values:

LI/6=21/7

Dividing on the right side of the equation:

LI/6=3

Solving for LI: Multiplying both sides of the equation by 6:

6(LI/6)=6(3)

LI=18


2.TV/VS=RU/US

Replacing the given values:

x/17.5=8/14

Simplifying the fraction on the right side of the equation: Dividing numerator and denominator by 2:

x/17.5=(8/2) / (14/2)

x/17.5=4/7

Solving for x: Multiplying both sides of the equation by 17.5:

17.5(x/17.5)=17.5(4/7)=(17.5/1)(4/7)

Multiplying:

x=(17.5 x 4) / (1 x 7)

x=70/7

Dividing:

x=10


3. BC is parallel to DE if AD/DB is equal to AE/EC:

AD/DB=15/12=(15/3) / (12/3)→AD/DB=5/4

AE/EC=10/8=(10/2) / (8/2)→AE/EC=5/4

Like AD/DB=5/4=AE/EC → BC is parallel to DE


4. BC is parallel to DE if AD/DB is equal to AE/EC:

AD/DB= 2DB / DB→AD/DB=2

AE/EC=30 / (AC-AE)=30 / (45-30)=30/15→AE/EC=2

Like AD/DB=2=AE/EC → BC is parallel to DE


5. If JH is a midsegment of triangle KLM:

x=(1/2)(30)

x=15


6. If JH is a midsegment of triangle KLM:

x=2(9)

x=18


7. If JH is a midsegment of triangle KLM: x=8


8. 2x+1=x+7

Solving for x. Grouping the x's on the left side of the equation: Subtracting x both sides of the equation:

2x+1-x=x+7-x

Subtracting:

x+1=7

Subtracting 1 both sides of the equation:

x+1-1=7-1

Subtracting:

x=6

2x+1=2(6)+1=12+1→2x+1=13

x+7=6+7→x+7=13


(3y-8)/(y+5)=(2x+1)/(x+7)

(3y-8)/(y+5)=13/13

(3y-8)/(y+5)=1

3y-8=y+5

Solving for y. Grouping the y's on the left side of the equation: Subtracting y both sides of the equation:

3y-8-y=y+5-y

Subtracting:

2y-8=5

Adding 8 both sides of the equation:

2y-8+8=5+8

Subtracting:

2y=13

Dividing both sides of the equation by 2:

2y/2=13/2

Dividing:

y=6.5



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3 years ago
Please help!! also please explain I would like to understand.​
ziro4ka [17]

Answer:

A

Step-by-step explanation:

First, put this equation in slope intercept form.

y=mx+b

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We can see that the point (2,1) is on the graph, and only Graph A has that point. Therefore, Graph A is the correct graph.

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