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Neporo4naja [7]
2 years ago
8

HERES THIS QUESTION PLEASE HELP

Mathematics
1 answer:
Oxana [17]2 years ago
7 0
27 hours in the maximum hours
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Joe has a bag containing 8 red sweets, 9 yellow ones and 11 green. He takes out a sweet and eats it, he takes out a second sweet
gizmo_the_mogwai [7]

Answer:

2/27

Step-by-step explanation:

This is a probability question and we are asked to estimate a particular probability. We proceed as follows:

The total number of sweets is given as 8 + 9+ 11= 28 sweets

He takes out a sweet to eat, the probability of this being a red sweet would be P(r1) = 8/28 = 2/7

Now he takes another sweet, we are asked to calculate the probability that this sweet is also red. Now after taking the first sweet, the number of sweets is now 27, while the number of red sweets is now 7. Hence the probability of having a red sweet taken in the second case would be p(r2) = 7/27

Now, the probability of both being red sweets can be calculated by multiplying both = 7/27 * 2/7 = 2/27

6 0
2 years ago
Tabitha has a coupon that will give her 30% off her total purchase at her favorite bath & candle store. If her total is $186
o-na [289]

Answer:

Step-by-step explanation:

186 devised by 30 is the answer

5 0
3 years ago
To find the value of the following expression,which operation should you do first and solve the problem following PEMDAS. The ex
Ymorist [56]
Add the 7 and 4, which =11.
multiply times the 5, =55
and 20-55= -35
8 0
3 years ago
There are 380 students at Cove Elementary School. The students voted on
Mrrafil [7]
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3 years ago
Consider f (x) = StartRoot x squared minus 1 EndRoot and g (x) = StartRoot x squared + 1 EndRoot. What value(s) of x would make
dalvyx [7]

Answer:

Any value of x

<em></em>

Step-by-step explanation:

Given

f(x) = \sqrt{x^2 - 1}

g(x) = \sqrt{x^2 + 1}

Required

What value of x is  f(g(x)) = g(f(x))

Solving for f(g(x))

f(x) = \sqrt{x^2 - 1}

f(g(x)) = \sqrt{(\sqrt{x^2 + 1})^2 - 1}

Solve the inner square

f(g(x)) = \sqrt{(x^2 + 1 - 1}

f(g(x)) = \sqrt{x^2 } }

f(g(x)) = x

Solving g(f(x))

g(x) = \sqrt{x^2 + 1}

g(f(x)) = \sqrt{(\sqrt{x^2 - 1})^2 + 1}

g(f(x)) = \sqrt{x^2 - 1 + 1}

g(f(x)) = \sqrt{x^2 }

g(f(x)) = x

Equate f(g(x)) and g(f(x))

f(g(x)) = g(f(x))

x = x

<em>This implies that </em>f(g(x)) = g(f(x))<em> at any value of x</em>

8 0
3 years ago
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