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Mars2501 [29]
2 years ago
8

A statement can help you organize the numbers to answer a question about percents

Mathematics
2 answers:
trasher [3.6K]2 years ago
8 0

Answer:

<h2>There was spent 27% of the total in clothing.</h2>

Step-by-step explanation:

According to the problem

Clothing $50.

Groceries $100.

Gas $35.

Total $185.

Now, to know the percentage spent on clothing, we just need to divide the amount of money spent on clothing by the total, then we multiply by 100 to express it in percentage.

\frac{50}{185} \times 100\% \approx 27\%

Therefore, there was spent 27% of the total in clothing.

In other words, $50 is 27% of $185.

Semenov [28]2 years ago
5 0

Answer:

<u> $50</u> is <u>27.02%</u> of <u>$185</u>.

Step-by-step explanation:

Money spent on clothing = $50

Money spent on Groceries = $100

Money spent on Gas = $35

Total money spent = $185

Our question is to find the percentage of total money that is spent on clothing.

Percentage money on clothing = \frac{Money spent on clothing}{Total money} × 100

                                                   = \frac{50}{185} × 100

                                                   = 27.02% (approx)

Hence,<u> $50</u> is <u>27.02%</u> of <u>$185</u>.  

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What would the equation be for d=0.68t + 1.5
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Step-by-step explanation:

a)

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d=0.68t + 1.5     if d depth is 4 ft    solve for t

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Answer:

a)

X    |  1        3      5       7

f(X) | 0.4   0.2   0.2    0.2

b) P(4

Step-by-step explanation:

For this case we have defined the cumulative distribution function like this:

F(X) = 0, x

F(X) = 0.4, 1 \leq x

F(X) = 0.6, 3 \leq x

F(X) = 0.8, 5 \leq x

F(X) = 1, x \geq 7

And we know that the general definition for the distribution function is given by:

F(x) = P(X \leq x) = \sum_{i\leq k} f(i)

Where f represent the density function.

Part a

For this case we need to find the density function, so we can find the values for the density for each value of X = 1,2,3,4,5,6,7,... since X is a discrete random variable.

f(1) = P(X=1) = P(X \leq 1) - P(X=0) = F(1) -F(0) = 0.4-0=0.4

f(2) = P(X=2) = P(X \leq 2) - P(X=0)- P(X=1) = F(2) -F(1) = 0.4-0.4=0

f(3) = P(X=3) = P(X \leq 3) - P(X=0)- P(X=1) -P(X=2) = F(3) -F(2) = 0.6-0.4=0.2

f(4) = P(X=4) = P(X \leq 4) - P(X=0)- P(X=1) -P(X=2)-P(X=3) = F(4) -F(3) = 0.6-0.6=0

f(5) = P(X=5) = P(X \leq 5) - P(X=0)- P(X=1) -P(X=2)-P(X=3)-P(X=4) = F(5) -F(4) = 0.8-0.6=0.2

f(6) = P(X=6) = P(X \leq 6) - P(X=0)- P(X=1) -P(X=2)-P(X=3)-P(X=4)-P(X=5) = F(6) -F(5) = 0.8-0.8=0

f(7) = P(X=7) = P(X \leq 7) - P(X=0)- P(X=1) -P(X=2)-P(X=3)-P(X=4)-P(X=5)-P(X=6) = F(7) -F(6) = 1-0.8=0.2

And for any value higher than 7 we have that:

x_i \in [8,9,10,...]

f(x_i) = F(X_i) -F(X_i -1) = 1-1=0

So then we have our density function defined like this:

X    |  1        3      5       7

f(X) | 0.4   0.2   0.2    0.2

Part b

For this case we want to find this probability P(4

And since the random variable is discrete we can write this like that:

P(4

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