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ValentinkaMS [17]
3 years ago
7

A model length of 12cm. Represents an actual length of 102 ft. What is the scale for the model

Mathematics
1 answer:
stealth61 [152]3 years ago
8 0
102 ft = 12 cm1224 in = 12 cm1 in = 0.00980 cm
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The sum of two numbers is -3. five times the first number equals 3 times the second number. find the two numbers
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-9/8 and -15/8. hope i helped

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2 years ago
9. The water level is 3 feet below your dock. The tide goes out, and the water level lowers 1 foot. A storm surge comes in, and
guapka [62]
We know that that the water level is 3 feet below your deck. (-3)
When the tide goes out, the water level lowers 1 foot. (-1)
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3 years ago
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In London today, four times the high temperature was more than twice the high temperature plus
4vir4ik [10]

Answer:

Let's define the high temperature as T.

We know that:

"four times T, was more than 2*T plus 66°C"

(i assume that the temperature is in °C)

We can write this inequality as:

4*T > 2*T + 66°C

Now we just need to solve this for T.

subtracting 2*T in both sides, we get:

4*T - 2*T > 2*T + 66°C - 2*T

2*T > 66°C

Now we can divide both sides by 2:

2*T/2 > 66°C/2

T > 33°C

So T was larger than 33°C

Notice that T = 33°C is not a solution of the inequality, then we should use the symbol ( for the set notation.

Then the range of possible temperatures is:

(33°C, ...)

Where we do not have an upper limit, so we could write this as:

(33°C, ∞°C)

(ignoring the fact that ∞°C is something impossible because it means infinite energy, but for the given problem it works)

4 0
3 years ago
Algebra 1 help! •giving brainliest!
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Compound interest:

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3 years ago
How many possible committees of 4 people can be chosen from 15 men and 20 women so that at least two women should be on each com
Mnenie [13.5K]

<span>If there has to be 2 men and 2 women, we know that we must take a group of 2 men out of the group of 15 men and a group of 2 women out of the group of 20 women. Therefore, we have:

(15 choose 2) x (20 choose 2)


(15 choose 2) = 105

(20 choose 2) = 190

190*105 = 19950

Therefore, there are 19950 ways to have a group of 4 with 2 men and 2women.</span>

 

<span>If there has to be 1 man and 3 women, we know that we must take a group of 1 man out of the group of 15 men and a group of 3 women out of the group of 20 women. Therefore, we have:

(15 choose 1) x (20 choose 3)


(15 choose 1) = 15

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Therefore, there are 17100 ways to have a group of 4 with 3 women and 1 man.</span>

 

<span>We now find the total outcomes of having a group with 4 women.

We know this is the same as saying (20 choose 4) = 4845</span>

Therefore, there are 4845 ways to have a group of 4 with 4 women.

 

We now add the outcomes of 2 women, 3 women, and 4 women and get the total ways that a committee can have at least 2 women.

 

19950 + 17100 + 4845 = 41895 ways that there will be at least 2 women in the committee


5 0
3 years ago
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