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Which of the following values cannot be probabilities? 0.04, 5 divided by 3, 1, 0, 3 divided by 5, StartRoot 2 EndRoot, negative 0.59, 1.49 Select all the values that cannot be probabilities. A. 1.49 B. 1 C. three fifths D. StartRoot 2 EndRoot E. five thirds F. 0 G. negative 0.59 H. 0.04

Answers

Answer 1

Answer:

Answer:

A. 1.49

D. √2

E. five thirds

G. - 0.59

Step-by-step explanation:

In order to be a probability, a value must be at least zero, or at most 1:

$0 \leq P\leq 1$

Evaluating each of the given values:

A. 1.49

1.49 is at least zero but it is greater than one, therefore 1.49 cannot be a probability.

B. 1

1 represents a probability of 100%, therefore this value can be a probability

C. three fifths

$0\leq (3)/(5) \leq 1$

Can be a probability

D. √2

$\sqrt 2 =1.41 > 1$

Cannot be a probability

E. five thirds

$(5)/(3)=1.67>1$

Cannot be a probability

F. 0

0 represents a probability of 0%, therefore this value can be a probability

G. - 0.59

Negative values cannot be probabilities.

H. 0.04

$0\leq 0.04 \leq 1$

Can be a probability

Answer 2

Answer:

Final answer:

Probabilities are values ranging from 0 to 1, inclusive. With this in mind, values 5/3, √2, -0.59, and 1.49 cannot be probabilities as they're either below 0 or above 1.

Explanation:

In the field of mathematics, specifically in statistics, a probability represents the likelihood of an event occurring and is always a value between 0 and 1, inclusively. The value 0 means that an event will not happen, whilst 1 means the event is certain to happen. Therefore, any value less than 0 or greater than 1 cannot be a probability.

Given the values: 0.04, 5 divided by 3, 1, 0, 3 divided by 5, √2, negative 0.59, and 1.49, the values that cannot be probabilities are:

Value 5 divided by 3 (which equals approximately 1.67)
Value √2 (which equals approximately 1.41)
Negative 0.59
1.49

These numbers do not lie within the range of 0 to 1, and hence, cannot represent probabilities.

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URGENT TEST I HAVE!!!!

Answers

Answer:

I think they are asking the area of the triangle so we are going with that.

Step-by-step explanation:

Easiest way to do this is find the area of the incomplete square, and then cut that in half for the triangle.

so it is 4 cm wide

and 8 cm tall

length x width = area

8 x 4 = 32

32/2 = 16

So your answer should be 16 cm

It took Fred 12 hours to travel over pack ice from one town to another town 360 miles away. During the return journey it took him 15 minutes

Answers

Step-by-step explanation:

what is your question please

At a certain car dealership, 20% of customers who bought a new vehicle bought an SUV, and 3% of them bought a black SUV (that is 3% of customers bought a vehicle that was an SUV and in black color). Given that a customer bought an SUV, what is the probability that it was black?

Answers

Answer: 0.15

Step-by-step explanation:

As per given , the probability that customers who bought a new vehicle bought an SUV : P(SUV) = 0.20

The probability that customer bought a vehicle that was an SUV and in black color : P(SUV and black) =0.03

Now by suing conditional probability formula,

If we have given that a customer bought an SUV, then the probability that it was black will be :

$\text{P(Black}|\text{SUV})=\frac{\text{P(SUV and Black)}}{\text{P(SUV)}}$

$=(0.03)/(0.20)=(3)/(20)=0.15$

Hence, the required probability is 0.15.

The probability that a customer who bought an SUV also bought a black SUV is 0.006, or 0.6% (expressed as a percentage).

To find the probability that a customer who bought an SUV also bought a black SUV, you can use conditional probability.

Let's define the following events:

A: A customer bought an SUV.

B: A customer bought a black SUV.

You are given that P(B|A) is the probability that a customer who bought an SUV also bought a black SUV, which is 3% or 0.03.

You want to find P(B|A), the probability that a customer who bought an SUV also bought a black SUV. You can use the following formula for conditional probability:

P(B|A) = (P(A and B)) / P(A)

Here, P(A and B) is the probability that a customer bought both an SUV and a black SUV, and P(A) is the probability that a customer bought an SUV.

You know that P(B|A) = 0.03 and P(A) = 0.20.

Now, you need to find P(A and B), the probability that a customer bought both an SUV and a black SUV. You can rearrange the formula:

P(A and B) = P(B|A) * P(A)

P(A and B) = 0.03 * 0.20

P(A and B) = 0.006

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(3x^2+2x-1)-(x^2-3x+4)

Answers

To evaluate the expression we shall have:
(3x^2+2x-1)-(x^2-3x+4)
putting like terms together we get:
(3x^2-x^2)+(2x+3x)+(-1-4)
=2x^2+5x-5
Answer: 2x^2+5x-5

Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie of this type contains at least two chocolate chips to be greater than 0.99. Find the smallest value of the mean that the distribution can take.

Answers

Answer:

$\lambda \geq 6.63835$

Step-by-step explanation:

The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".

Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that $X \sim Poisson(\lambda)$

The probability mass function for the random variable is given by:

$f(x)=(e^(-\lambda) \lambda^x)/(x!) , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda$

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

$P(X\geq 2)=1-P(X<2)=1-P(X\leq 1)=1-[P(X=0)+P(X=1)]$

Using the pmf we can find the individual probabilities like this:

$P(X=0)=(e^(-\lambda) \lambda^0)/(0!)=e^(-\lambda)$

$P(X=1)=(e^(-\lambda) \lambda^1)/(1!)=\lambda e^(-\lambda)$

And replacing we have this:

$P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^(-\lambda) +\lambda e^(-\lambda)[]$

$P(X\geq 2)=1-e^(-\lambda)(1+\lambda)$

And we want this probability that at least of 99%, so we can set upt the following inequality:

$P(X\geq 2)=1-e^(-\lambda)(1+\lambda)\geq 0.99$

And now we can solve for $\lambda$

$0.01 \geq e^(-\lambda)(1+\lambda)$

Applying natural log on both sides we have:

$ln(0.01) \geq ln(e^(-\lambda)+ln(1+\lambda)$

$ln(0.01) \geq -\lambda+ln(1+\lambda)$

$\lambda-ln(1+\lambda)+ln(0.01) \geq 0$

Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.

Using the Newthon Raphson method, we apply this formula:

$x_(n+1)=x_n -(f(x_n))/(f'(x_n))$

Where :

$f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)$

$f'(x_n)=1-(1)/(1+\lambda)$

Iterating as shown on the figure attached we find a final solution given by:

$\lambda \geq 6.63835$

Final answer:

The problem pertains to Poisson Distribution in probability theory, focusing on finding the smallest mean (λ) such that the probability of having at least two chocolate chips in a cookie is more than 0.99. This involves solving an inequality using the formula for Poisson Distribution.

Explanation:

This problem pertains to the Poisson Distribution, often used in probability theory. In particular, we're looking at the number of events (in this case, the number of chocolate chips) that occur within a fixed interval. Here, the interval under study is a single cookie. The question requires us to find the smallest value of λ (the mean value of the distribution) such that the probability of getting at least two chocolate chips in a cookie is more than 0.99.

Using the formula for Poisson Distribution, the probability of finding k copies of an event is given by:

P(X=k) = λ^k * exp(-λ) / k!

The condition here is that the probability of finding at least 2 copies is more than 0.99. Therefore, you formally need to solve the inequality:

P(X>=2) = 1 - P(X=0) - P(X=1) > 0.99

Substituting the values of P(X=0) and P(X=1) from our standard formula, you will need to calculate and find the smallest value of λ that satisfies this inequality.

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Janice is studying the effect of pipe size on water flow rates. In12 minutes, 463.2 gallons of water flowed from a 2-inch pipe. In
7 minutes, 1730.4 gallons of water flowed from a 4-inch pipe.
Based on Janice's data, what is the difference in flow rate
between a 2-inch and 4-inch pipe?

Answers

Answer:

208.6 gal/min

Step-by-step explanation:

For 2" pipe,

Given Volume = 463.2 gal, time = 12 min

flow rate for 2" pipe

= Volume ÷ time

= 463.2÷12

= 38.6 gal/min

For 4" pipe,

Given Volume = 1730.4 gal, time = 7 min

flow rate for 4" pipe

= Volume ÷ time

= 1730.4÷7

= 247.2 gal/min

Difference in flow rate = 247.2 - 38.6 = 208.6 gal/min

Final answer:

The difference in flow rate between a 2-inch and 4-inch pipe, based on Janice's data, is 208.6 gallons per minute.

Explanation:

To determine the difference in flow rate between a 2-inch and a 4-inch pipe, we first need to calculate the flow rate for each pipe. This can be done by dividing the amount of water that flowed within a given time by that time.

For the 2-inch pipe: 463.2 gallons flowed in 12 minutes, so the flow rate is 463.2 / 12 = 38.6 gallons per minute.

For the 4-inch pipe: 1730.4 gallons flowed in 7 minutes, so the flow rate is 1730.4 / 7 = 247.2 gallons per minute.

Now, to find the difference in flow rate between the two pipes, we subtract the smaller flow rate from the larger one. Thus, 247.2 - 38.6 = 208.6 gallons per minute.

Therefore, the difference in flow rate between a 2-inch and 4-inch pipe, based on Janice's data, is 208.6 gallons per minute.

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