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3 years ago

Which of the following are integer solutions to the inequality below? −2≤×<3

Mathematics

1 answer:

Phoenix [80]3 years ago

6 0

Answer:

-2, - 1, 0, 1, 2

Step-by-step explanation:

The greater than or equal to sign (≤) demonstrates that the unknown is equal to -2 and greater, but is less than 3, so the largest integer solution is 2, not 3

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Owen’s bedroom has a perimeter of 46 feet. If the length of the bedroom is 11 feet, what is the width

AnnyKZ [126]

Perimeter of a rectangle is given by (2 × Length(l)) + (2 × Width(w)) = 2l + 2w

l= 11 feet

2(11) + 2w = 46

2w = 46 - 22 = 24

w = 24/2 = 12 feet

Width = 12 feet.

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Step-by-step explanation:

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3 years ago

Let A and B be events with =PA0.7, =PB0.3, and =PA or B0.9. (a) Compute PA and B. (b) Are A and B mutually exclusive? Explain. (

kvasek [131]

Answer:

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

And if we solve for $P(A \cap B)$ we got:

$P(A \cap B) = P(A) + P(B) -P(A\cup B)= 0.7+0.3-0.9 = 0.1$

b) False

The reason is because we don't satisfy the following relationship:

$P(A\cup B) = P(A) + P(B)$

We have that:

$0.9 \neq 0.3+0.7 =1$

c) False

In order to satisfy independence we need to have the following condition:

$P(A \cap B) = P(A) *P(B)$

And for this case we don't satisfy this relation since:

$0.1 \neq 0.7*0.3 = 0.21$

Step-by-step explanation:

For this case we have the following probabilities given:

$P(A) = 0.7, P(B) =0.7, P(A \cup B) =0.9$

Part a

We want to calculate the following probability: $P(A \cap B)$

And we can use the total probability rule given by:

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

And if we solve for $P(A \cap B)$ we got:

$P(A \cap B) = P(A) + P(B) -P(A\cup B)= 0.7+0.3-0.9 = 0.1$

Part b

False

The reason is because we don't satisfy the following relationship:

$P(A\cup B) = P(A) + P(B)$

We have that:

$0.9 \neq 0.3+0.7 =1$

Part c

False

In order to satisfy independence we need to have the following condition:

$P(A \cap B) = P(A) *P(B)$

And for this case we don't satisfy this relation since:

$0.1 \neq 0.7*0.3 = 0.21$

4 0

3 years ago

Which of the following is not a property of a chi-square distribution?

laiz [17]

Answer:

c) Is not a property (hence (d) is not either)

Step-by-step explanation:

Remember that the chi square distribution with k degrees of freedom has this formula

$\chi_k^2 = \matchal{N}_1^2 + \matchal{N}_2^2 + ... + \, \matchal{N}_{k-1}^2 + \matchal{N}_k^2$

Where N₁ , N₂m .... $N_k$ are independent random variables with standard normal distribution. Since it is a sum of squares, then the chi square distribution cant take negative values, thus (c) is not true as property. Therefore, (d) cant be true either.

Since the chi square is a sum of squares of a symmetrical random variable, it is skewed to the right (values with big absolute value, either positive or negative, will represent a big weight for the graph that is not compensated with values near 0). This shows that (a) is true

The more degrees of freedom the chi square has, the less skewed to the right it is, up to the point of being almost symmetrical for high values of k. In fact, the Central Limit Theorem states that a chi sqare with n degrees of freedom, with n big, will have a distribution approximate to a Normal distribution, therefore, it is not very skewed for high values of n. As a conclusion, the shape of the distribution changes when the degrees of freedom increase, because the distribution is more symmetrical the higher the degrees of freedom are. Thus, (b) is true.

6 0

3 years ago