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kirza4 [7]
3 years ago
7

Add.

Mathematics
2 answers:
gayaneshka [121]3 years ago
7 0

Answer:

8 7/24 hope this helps

Step-by-step explanation:

S_A_V [24]3 years ago
6 0
I really don’t understand the question
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Simplify and solve for the unknown. Use order of operations as needed. Check your work.
antiseptic1488 [7]

Answer:

5

Step-by-step explanation:

Given: 5^3– 10^2 = X(8 – 2) + 2X – 3X

Solving the equation to find the value of x.

⇒ 5^{3} -10^{2} = x(8-2)+2x-3x

Using the distributive property of multiplication, multiplying x with 8 and -2

⇒125-100= 8x-2x+2x-3x

cancelling out -2x and 2x

⇒ 25= 8x-3x

⇒ 25= 5x

Multiplying both side by 5

⇒ x= \frac{25}{5}

∴ x= 5

Hence, the value of unknown is 5.

3 0
3 years ago
The question for this problem is “The owner of a food cart sells an average of 120 frozen treats per day during the summer”.
ziro4ka [17]

Answer:

i think its b

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
How many times does the minute hand on the clock pass 6
julsineya [31]

Answer:

12 times a day.

Step-by-step explanation:

The minute hand will pass 6 on a clock from midnight to noon each day for 12 times a day.

3 0
3 years ago
Read 2 more answers
How many times does 81,000 goes into 8,000
Effectus [21]
Zero. That number cannot go into that one
8 0
3 years ago
In a certain community, 36 percent of the families own a dog and 22 percent of the families that own a dog also own a cat. In ad
zysi [14]

Answer:  a) 0.0792   b) 0.264

Step-by-step explanation:

Let Event D = Families own a dog .

Event C = families own a cat .

Given : Probability that families own a dog : P(D)=0.36

Probability that families own a dog also own a cat : P(C|D)=0.22

Probability that families own a cat : P(C)= 0.30

a) Formula to find conditional probability :

P(B|A)=\dfrac{P(A\cap B)}{P(A)}\\\\\Rightarrow P(A\cap B)=P(B|A)\times P(A)   (1)

Similarly ,

P(C\cap D)=P(C|D)\times P(D)\\\\=0.22\times0.36=0.0792

Hence, the probability that a randomly selected family owns both a dog and a cat : 0.0792

b) Again, using (2)

P(D|C)=\dfrac{P(C\cap D)}{P(C)}\\\\=\dfrac{0.0792}{0.30}=0.264

Hence, the conditional probability that a randomly selected family owns a dog given that it owns a cat = 0.264

5 0
3 years ago
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