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Anna007 [38]
3 years ago
10

Plz help me well mark brainliest if correct!!.. .

Mathematics
1 answer:
Mashutka [201]3 years ago
6 0

Answer:

70 ft

Step-by-step explanation:

20 + 20 + 15 + 15 = 70

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3/4=j-1/2 solve the equation check your solution
DIA [1.3K]
3/4= j-1/2
first to add 1/2 to both sides to solve for j 
3/4 + 1/2= j
second you find a common denominator in order to add 3/4 and 1/2, the common denominator would be 4, you would have to change 3/4 at all because the denominator is already 4 but in order to make the denominator of 1/2, 4 you would have to multiply both the denominator and numerator by 2 so 
3/4 + 2/4 = 5/4
so 5/4 =j

to check your answer you plug 5/4 in for j 
3/4 = 5/4 -1/2
again you need to find a common denominator between 5/4 and 1/2 which again would be 4, and again you wouldn't change 5/4 but you would multiply both the numerator and the denominator of 1/2 so 
3/4= 5/4-2/4
5-2= 3 and you would keep the 4 so 
5/4 - 2/4 = 3/4

so j =  5/4
5 0
3 years ago
Jeff uses 3 fifth-size strips to mode 3/5. He wants to use tenth-size strips to model an equivalent fraction. How many tenth-siz
galben [10]
Wouldn't it just be 6 tenth sized for 6/10
4 0
4 years ago
diana bnrought a car for 5,000 and plan to sell it to make a 20% profit .what should she sell it for to make the profit she desi
Vikentia [17]
$6,000


5,000•1.2=$6,000
4 0
3 years ago
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}

7 0
3 years ago
Giving Brainliest to whoever answers well &amp; first. 
Ne4ueva [31]
Hopefully this helped :D

7 0
3 years ago
Read 2 more answers
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