Answer: Choice B. Dependent; probabilities are not equal
P(picky) = 34%
P(picky | cat) = 44%
This is a vertical bar separating "picky" and "cat". It is not an uppercase letter i or lowercase letter L.
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Explanation:
There are 17 total picky eaters out of 50 pets total. The probability of randomly choosing a picky eater is
P(picky) = 17/50 = 0.34 = 34%
The notation P(picky | cat) translates to "probability of getting a picky eater given the pet is a cat". In other words, we have prior knowledge the pet is a cat. So we only focus on the cat row. We have 11 picky cats out of 25 cats total. This means
P(picky | cat) = 11/25 = 0.44 = 44%
Notice how this result is different from 34% found earlier. The fact that we have prior knowledge the pet selected is a cat changes the probability of P(picky). This means the events are dependent. If they were independent events, then P(picky) = P(picky | cat) would be true.
Not sure about the second part but
2a+152=252
subtract 152 from both sdies
2a=100
divide both sdies by 2
a=50
Answer:
The correct options is;
Midsegment of a Triangle Theorem
Step-by-step explanation:
The Midsegment of a Triangle Theorem states that when there is a segment that joins the midpoints of two adjacent sides of a triangle, that segment will be half the length of the third side of the triangle and parallel to it (the third side of the triangle)
Therefore, each triangle can have at least, three midsegments constructed to be parallel to each of the three sides
m∠ADE = 36° (Given)
m∠DAE = 90° (Definition of a right angle)
m∠AED = 54° (Triangle Sum Theorem)
Segment DE joins the midpoints of segments AB and AC (Given)
Segment DE is parallel to segment BC (<u>Midsegment of a Triangle Theorem</u>)
∠ECB ≅ ∠AED (Corresponding angles are congruent)
∴ ∠ECB = 54° (Substitution property).
Answer:
- see the attachment for a graph
- yes, you can carry 5 math books in one load (along with 0–2 science books)
Step-by-step explanation:
If x and y represent the number of math and science books you're carrying, respectively, then 3x and 4y represent their weights in pounds.
The total weight of the carry will be 3x+4y, and you want that to be at most 24 pounds. The expression modeling this is ...
... 3x +4y ≤ 24
A graph of this inequality is shown in the attachment. (We have added the constraints that the number of books not be negative.)
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5 math books will weigh 5·3 = 15 pounds, so will be within the limit you can carry.
Just here for points sorry