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11 months ago

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A state's recidivism rate is 22%. This means about 22% of released prisoners end up back in prison (within three years). Su

ppose two randomly selected prisoners who have been released are studied. What is the probability that both of them go back to prison?

1 answer:

storchak [24]11 months ago

7 0

The probability that both of them go back to prison is 4.48%.

<h3>What is probability?</h3>

Probability is a measure of the likelihood of an event occurring.
Many events cannot be predicted with total certainty.
We can predict only the chance of an event to occur i.e., how likely they are going to happen, using it.
The probability of an event is a number between 0 and 1, where, roughly speaking, 0 indicates the impossibility of the event and 1 indicates certainty.

The state recidivism rate is given by 22%.

So, the probability that both of the prisoners will go back to prison will be:

= 22% × 22%
= 0.22 × 0.22
= 4.84%

Hence, the probability that both of them go back to prison is 4.48%.

To know more about probability, visit:

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Write a compound inequality that represents the following scenario:

Inessa05 [86]

The compound inequality that represents the two following scenarios are:

65 < f ≤ 4
8 ≤ f ≤ 12

A compound inequality usually puts together two or more simple inequalities statements together.

Following the assumption from the given information that;

a free single scoop cone = f

<h3>1.</h3>

The age group of individuals designated to receive the free single scoop cones is:

people who are older than 65 i.e. > 65
children that are 4 or under 4 i.e. ≤ 4

Thus, the compound inequality that is appropriate to express both conditions is:

65 < f ≤ 4

<h3>2.</h3>

On Tuesdays, the least amount of flavors = 8
The addition amount of extra flavors they can add = 4

Now, we can infer that the total amount of flavors = 8 + 4 = 12

Thus, the compound inequality that is appropriate to express both conditions is:

Least amount of flavors ≤ f ≤ total amount of flavors
8 ≤ f ≤ 12

Therefore, we can conclude that the compound inequality that represents the two following scenarios are:

65 < f ≤ 4
8 ≤ f ≤ 12

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1 year ago

List one pair of vertical angles.

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Answer:

1 and 4, 2 and 3, 5 and 8, 6 and 7

Step-by-step explanation:

Any of those would work

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2 years ago

Carlton is using the following data samples to make a claim about the house values in his neighborhood:

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The answer is A because there is a outlier which is $100,000

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2 years ago

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8x2y2 – + 4x4 – 7xy3

Elodia [21]

<span>a
-7xy3 + 8x2y2 - + 4x4</span>

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2 years ago

Which series of transformations will not map figure H onto itself

7nadin3 [17]

Answer:

D

Step-by-step explanation:

Given a square with vertices at points (2,1), (1,2), (2,3) and (3,2).

Consider option A.

1st transformation $(x+0,y-2)$ will map vertices of the square into points

$(2,1)\rightarrow (2,-1);$
$(1,2)\rightarrow (1,0);$
$(2,3)\rightarrow (2,1);$
$(3,2)\rightarrow (3,0).$

2nd transformation = reflection over y = 1 has the rule (x,2-y). So,

$(2,-1)\rightarrow (2,3);$
$(1,0)\rightarrow (1,2);$
$(2,1)\rightarrow (2,1);$
$(3,0)\rightarrow (3,2)$

These points are exactly the vertices of the initial square.

Consider option B.

1st transformation $(x+2,y-0)$ will map vertices of the square into points

$(2,1)\rightarrow (4,1);$
$(1,2)\rightarrow (3,2);$
$(2,3)\rightarrow (4,3);$
$(3,2)\rightarrow (5,2).$

2nd transformation = reflection over x = 3 has the rule (6-x,y). So,

$(4,1)\rightarrow (2,1);$
$(3,2)\rightarrow (3,2);$
$(4,3)\rightarrow (2,3);$
$(5,2)\rightarrow (1,2)$

These points are exactly the vertices of the initial square.

Consider option C.

1st transformation $(x+3,y+3)$ will map vertices of the square into points

$(2,1)\rightarrow (5,4);$
$(1,2)\rightarrow (4,5);$
$(2,3)\rightarrow (5,6);$
$(3,2)\rightarrow (6,5).$

2nd transformation = reflection over y = -x + 7 will map vertices into points

$(5,4)\rightarrow (3,2);$
$(4,5)\rightarrow (2,3);$
$(5,6)\rightarrow (1,2);$
$(6,5)\rightarrow (2,1)$

These points are exactly the vertices of the initial square.

Consider option D.

1st transformation $(x-3,y-3)$ will map vertices of the square into points

$(2,1)\rightarrow (-1,-2);$
$(1,2)\rightarrow (-2,-1);$
$(2,3)\rightarrow (-1,0);$
$(3,2)\rightarrow (0,-1).$

2nd transformation = reflection over y = -x + 2 will map vertices into points

$(-1,-2)\rightarrow (4,3);$
$(-2,-1)\rightarrow (3,4);$
$(-1,0)\rightarrow (2,3);$
$(0,-1)\rightarrow (3,2)$

These points are not the vertices of the initial square.

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3 years ago