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2 years ago

If cyanide in a stream next to a gold mine increases from 240 ppm to 360 ppm, what percent increase is this?

Mathematics

1 answer:

avanturin [10]2 years ago

4 0

Given :

Initial concentration , 240 ppm .

Final concentration , 360 ppm .

To Find :

Percent increase.

Solution :

Percentage increase is given by :

$=\dfrac{Final-Initial}{Initial}\times 100\\\\=\dfrac{360-240}{240}\times 100\\\\=50\%$

Therefore , percent increase is 50 % .

Hence , this is the required solution .

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What is the area of the composite figure whose vertices have the following coordinates? (−2, −2) , (3, −2) , (5, −4) , (1, −8) ,

svet-max [94.6K]

Area of a rectangle = W X L
Area of a rectangle = 5 X 2
Area of a rectangle = 10

Area of triangle 1= 1/2 X B X H
Area of triangle 1= 1/2 X 2 X 2
Area of triangle 1= 1/2 X 4
Area of triangle 1= 2

Area of triangle 2= 1/2 X B X H
Area of triangle 2= 1/2 X 7 X 4
Area of triangle 2= 1/2 X 28
Area of triangle 2= 14

Area of a rectangle + Area of triangle 1 + Area of triangle 2=

10 + 2 + 14 = 36

7 0

2 years ago

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Round all answers to the nearest whole number. How many of the people surveyed make less than 5 calls per day? What percentage o

dimaraw [331]

Answer:

a). 18

b). 26%

c). 39

Step-by-step explanation:

Given question is incomplete; here is the complete question attached.

a). In this part we have to calculate the number of the people surveyed who make less than 5 calls.

From the given table,

Number of people surveyed who make less than 5 calls Or 1 - 4 calls = 18

b). Total number of people surveyed = 18 + 11 + 5 + 3 + 2 = 39

Number of people surveyed, make at least 9 calls = 5 + 3 + 2 = 10

Percentage of these people = $\frac{\text{People who make calls more than 9 calls per day}}{\text{Total number of people surveyed}}\times 100$

= $\frac{10}{39}\times 100$

= 25.64%

≈ 26%

c). Total number of people surveyed = 18 + 11 + 5 + 3 + 2 = 39

7 0

2 years ago

8. Which of the numbers in each pair is farther to the left on the number line?

galben [10]

A. 305
b. 900
c. 46
d. 157,019

6 0

2 years ago

The Big River Casino is advertising a new digital lottery-style game called Instant Lotto. The player can win the following mone

zysi [14]

Answer:

(a) The expected value of the prize for one play of Instant Lotto is $3.50.

(b) The probability that the visitor wins some prize at least twice in the 20 free plays is 0.2641.

Step-by-step explanation:

(a)

The probability distribution of the monetary prizes that can be won at the game called Instant Lotto is:

X P (X = x)

$10 0.05

$15 0.04

$30 0.03

$50 0.01

$1000 0.001

$0 0.869

___________

Total = 1.000

Compute the expected value of the prize for one play of Instant Lotto as follows:

$E(X)=\sum x\cdot P (X=x)$

$=(10\times 0.05)+(15\times 0.04)+(30\times 0.03) \\+ (50\times 0.01)+(1000\times 0.001)+(0\times 0.869)\\=0.5+0.6+0.9+0.5+1+0\\=3.5$

Thus, the expected value of the prize for one play of Instant Lotto is $3.50.

(b)

Let X = number of times a visitor wins some prize.

A visitor to the casino is given n = 20 free plays of Instant Lotto.

The probability that a visitor wins at any of the 20 free plays is, p = 1/20 = 0.05.

The event of a visitor winning at a random free play is independent of the others.

The random variable X follows Binomial distribution with parameters n = 20 and p = 0.05.

Compute the probability that the visitor wins some prize at least twice in the 20 free plays as follows:

P (X ≥ 2) = 1 - P (X < 2)

= 1 - P (X = 0) - P (X = 1)

$=1-[{20\choose 0}0.05^{0}(1-0.05)^{20-0}]-[{20\choose 1}0.05^{1}(1-0.05)^{20-1}]\\=1-0.3585-0.3774\\=0.2641$

Thus, the probability that the visitor wins some prize at least twice in the 20 free plays is 0.2641.

(c)

Let X = number of people who play Instant Lotto each day.

The random variable X is normally distributed with a mean, μ = 800 people and a standard deviation, μ = 310 people.

Compute the probability that a randomly selected day has at least 1000 people play Instant Lotto as follows:

Apply continuity correction:

P (X ≥ 1000) = P (X > 1000 + 0.50)

= P (X > 1000.50)

$=P(\frac{X-\mu}{\sigma}>\frac{1000.50-800}{310})$

$=P(Z>0.65)\\=1-P(Z$

Thus, the probability that a randomly selected day has at least 1000 people play Instant Lotto is 0.2579.

5 0

2 years ago

Alice studies the relationship between climate and heart disease around the world. H(t)H(t)H, left parenthesis, t, right parenth

ddd [48]

Answer:

The domain of the function H(t), is [-5, 30].

The range of the function H(t), is [(10% + average), (average - 20%)]

Step-by-step explanation:

The domain of a function is the complete set of possible values of the independent variable.

For this question, the function is H(t), with the temperature, t, serving as the independent variables and H(t) the evidently dependent variable.

The domain of a function refers to all the possible independent variable values that will give corresponding real dependent variable values.

For this question, Alice's model has the probability for the occurrence of heart disease (in percents relative to the global average) at an area, H(t) varying with the temperature of that area in degree Celsius.

At a temperature of -5°C (the lowest temperature in the model), the probability is 10% above the average.

Then, the probability decreases with increase in temperature, taking a value 20% lower than the average when the temperature is at its highest of 30°C in the model.

So, temperature ranges from -5°C to 30°C and the probability for the occurrence of heart disease ranges from 10% above the average to 20% below the average.

The domain of the function H(t), from the definition given above would therefore be [-5, 30]

And the range of the function H(t), is [(10% + average), (average - 20%)]

Hope this Helps!!!

7 0

2 years ago

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