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Irina-Kira [14]
3 years ago
8

If the Probability of an event happening is 70%, what are the ODDS of the event happening?

Mathematics
1 answer:
elena55 [62]3 years ago
7 0

Answer:

a) 7:10

Step-by-step explanation:

An example of this would be having seven blue marbles and three red marbles in a jar.  If you randomly select one, the chance of that marble being blue is seventy, because 10 (total number of marbles) divided by seven (number of blue marbles) is 0.70, and if you convert this to a whole number (by multiplying it by 100) you get 70.

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Please i need some help on this question
Pachacha [2.7K]

Answer:

1) For 3 tickets, the cost is $12

2) For 5 tickets, the cost is $20

3) $28 can purchase 7 tickets

4) For 10 tickets, the cost is $40

Step-by-step explanation:

1) For 3 tickets

8 = 2

x = 3

(8 ÷ x) = (2 ÷ 3)

Cross multiply,

2x = 8 x 3

2x = 24

x = 24 ÷ 2

x = $12

Use the same principle for the other questions. Use X (or any other variable) to represent the missing value, I e. the value you are calculating.

I hope this helps.

7 0
3 years ago
Solve equation: 15=2y - 5(quickly please)
andrew11 [14]


15+5=2y-5+5

=20 = 2y

20/2 = 2y/2

10 = y

5 0
3 years ago
What is the coefficient of the term x7y in the expansion of (x + y)^8?
s344n2d4d5 [400]
<span>The coefficient for x^7*y is equal to 8Choose7 which is 8!/7! which is 8, or answer b. Pascal's triangle may be used to capture this as well, or use the exponent as above.</span>
3 0
4 years ago
Joel made some muffins he gave 1/4 of the muffins to a neighbor. He took 3/8 of the muffins to school. What fraction of the muff
geniusboy [140]
3/8 of the muffin is left
4 0
4 years ago
Evaluate the integral ∫2−1|x−1|dx
defon

I think you might be referring to the definite integral,

\displaystyle \int_{-1}^2|x-1|\,\mathrm dx

Recall the definition of absolute value:

|x| = \begin{cases}x&\text{if }x\ge0\\-x&\text{if }x

Then |x-1|=x-1 if x\ge1, and |x-1|=1-x is x. So spliting up the integral at <em>x</em> = 1, we have

\displaystyle \int_{-1}^2|x-1|\,\mathrm dx = \int_{-1}^1(1-x)\,\mathrm dx + \int_1^2(x-1)\,\mathrm dx

The rest is simple:

\displaystyle \int_{-1}^2|x-1|\,\mathrm dx = \left(x-\dfrac{x^2}2\right)\bigg|_{-1}^1 + \left(\dfrac{x^2}2-x\right)\bigg|_1^2 \\\\ = \left(\left(1-\frac12\right)-\left(-1-\frac12\right)\right) + \left(\left(2-2\right)-\left(\frac12-1\right)\right) \\\\ = \boxed{\frac52}

5 0
3 years ago
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