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astraxan [27]
3 years ago
5

Different shapes are drawn on cards and then the cards are placed in a bag. The number of cards for each shape is shown in the t

able. Shape Number of cards rectangle:32 hexagon:20 triangle:48 What is the probability that a randomly selected card has a hexagon drawn on it?
Mathematics
1 answer:
Verdich [7]3 years ago
8 0
This question was posted almost a week ago, and I am not sure if you still need the answer, but I'll explain it.


The total number of all of the shapes were placed in the bag are equal to 100. The number comes from adding 32+20+48=100.
This is our total number and will be the denominator.
There are a total of 32 cards which have a hexagon, this number will be our numerator.

So far we have 32/100 , but this is not simplest form.

Next to reduce the fraction, the GCF must be found.

32: 1, 2, 4, 8, 32
100: :
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In which number does one digit 6 have a value of 1 /100 of the other 6?
maw [93]

Answer:

The answer is C.

Step-by-step explanation:

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3 years ago
Directions:Solve this equation
avanturin [10]

Answer:

-3.21538461538

Step-by-step explanation:

\mathrm{Equation\;Given:}\; 5+5x=\frac{12x-\frac{12}{5}\cdot \:7}{5}

\mathrm{Multiply\;both\;sides\;by\;5}:   5\cdot \:5+5x\cdot \:5=\frac{12x-\frac{12}{5}\cdot \:7}{5}\cdot \:5

\mathrm{Now\;we\;have\;}:  25+25x=12x-\frac{84}{5}

\mathrm{Subtract\;by\;sides\;by\;25}:   25+25x-25=12x-\frac{84}{5}-25

\mathrm{Thus\;we\;have}:  25x=12x-\frac{209}{5}

\mathrm{Subtracting\;12x\;from\;both\;sides}:  25x-12x=12x-\frac{209}{5}-12x

\mathrm{So\;then\;we\;have:} 13x=-\frac{209}{5}

\mathrm{Now\;divide\;both\;sides\;by\;13}:  \frac{13x}{13}=\frac{-\frac{209}{5}}{13}

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\mathrm{Decimal\;form}:-3.21538461538

<u><em>Kavinsky</em></u>

7 0
2 years ago
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solong [7]

Given:

P(A)=0.4, P(B)=0.7, P(A\text{ and }B)=0.3

To find:

The value of P(A\text{ or }B).

Solution:

We know that,

P(A\text{ and }B)=P(A\cap B)

P(A\text{ or }B)=P(A\cup B)

It means, we have P(A)=0.4, P(B)=0.7, P(A\cap B)=0.3 and we need to find the value of  P(A\cup B).

P(A\cup B)=P(A)+P(B)-P(A\cap B)

P(A\cup B)=0.4+0.7-0.3

P(A\cup B)=1.1-0.3

P(A\cup B)=0.8

Therefore, the value of P(A\text{ or }B) is 0.8.

4 0
3 years ago
If x is a real number such that x^3=216, then what does x^2+x=?
BartSMP [9]
\sqrt[3]{216} =6 then
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