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padilas [110]
3 years ago
10

If V/7 =8 what is the value of V

Mathematics
2 answers:
ZanzabumX [31]3 years ago
5 0

Steps to solve:

v/7 = 8

~Simplify

1/7v = 8

~Multiply 7 to both sides

v = 56

Best of Luck!

Vitek1552 [10]3 years ago
4 0

Answer:

v=56

Step-by-step explanation:

multiply bothe sides of the equation by 7

7x v/7= 7x8

simplify

v=7x8

therefore v=56

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What is the length of a segment with endpoints (2, 5) and (10, 11) *
yuradex [85]

Answer: 14 Units

Step-by-step explanation: Subtract (10-2) and (11-5)

4 0
3 years ago
At a certain pizza parlor, % of the customers order a pizza containing onions, % of the customer's order a pizza containing saus
wel

Answer:

At a certain pizza parlor,36 % of the customers order a pizza containing onions,35 % of the customers order a pizza containing sausage, and 66% order a pizza containing onions or sausage (or both). Find the probability that a customer chosen at random will order a pizza containing both onions and sausage.

Step-by-step explanation:

Hello!

You have the following possible pizza orders:

Onion ⇒ P(on)= 0.36

Sausage ⇒ P(sa)= 0.35

Onions and Sausages ⇒ P(on∪sa)= 0.66

The events "onion" and "sausage" are not mutually exclusive, since you can order a pizza with both toppings.

If two events are not mutually exclusive, you know that:

P(A∪B)= P(A)+P(B)-P(A∩B)

Using the given information you can use that property to calculate the probability of a customer ordering a pizza with onions and sausage:

P(on∪sa)= P(on)+P(sa)-P(on∩sa)

P(on∪sa)+P(on∩sa)= P(on)+P(sa)

P(on∩sa)= P(on)+P(sa)-P(on∪sa)

P(on∩sa)= 0.36+0.35-0.66= 0.05

I hope it helps!

5 0
3 years ago
What is the probability of rolling a number greater than or equal to 9 with the sum of two dice, given that at least one of the
olga55 [171]
Conditional probablility P(A/B) = P(A and B) / P(B). Here, A is sum of two dice being greater than or equal to 9 and B is at least one of the dice showing 6. Number of ways two dice faces can sum up to 9 = (3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6) = 10 ways. Number of ways that at least one of the dice must show 6 = (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6), (6, 5), (6, 4), (6, 3), (6, 2), (6, 1) = 11 ways. Number of ways of rolling a number greater than or equal to 9 and at least one of the dice showing 6 = (3, 6), (4, 6), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6) = 7 ways. Probability of rolling a number greater than or equal to 9 given that at least one of the dice must show a 6 = 7 / 11
6 0
3 years ago
I need help with this
Harrizon [31]

105 \div 8.7

4 0
3 years ago
Read 2 more answers
How can division patterns help us divide multiples of 10
KengaRu [80]
Division using multiples of 10 is different than how most of us learned how to divide. <span>The idea of multiple is what number can 10 go into without a remainder. That is easy. Ten ends in a zero. Thus 10 goes into numbers ending in zero. An example is 60. Ten ends in a zero; 60 ends in a zero. It will divide evenly. </span>
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3 years ago
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