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GaryK [48]
3 years ago
7

Jake can carry 6 1/4 pounds of wood in from the barn. His father can carry 15/7 times as much as Jake. How many pounds can Jakes

father carry?
Mathematics
1 answer:
bazaltina [42]3 years ago
4 0
If you would like to know how many pounds can Jake's father carry, you can calculate this using the following steps:

6 1/4 = 25/4 pounds of wood

15/7 * 6 1/4 = 15/7 * 25/4 = 375/28 = 13 11/28 pounds

Result: Jake's father can carry 13 11/28 pounds of wood.
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Given Point A (3,-4) and Point B (8,6) on directed line segment AB, what is the y-coordinate of Point F
Allisa [31]

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(6,2) y-coordinate is 2

Step-by-step explanation:

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3 years ago
Jack has some marbles. He has ten more red marbles than blue marbles. He has fewer yellow marbles than blue marbles.
kap26 [50]

Answer:

<u>Jack has at least 15 marbles: 12 red, 2 blue and 1 yellow.</u>

Step-by-step explanation:

1. Let's review the information given to us to solve the problem correctly:

Number of red marbles = 10 more than blue marbles

Number of yellow marbles fewer than blue marbles

2. How many of each color marble could Jack have?

r = Number of red marbles

b = Number of blue marbles

y = Number of yellow marbles

r = b + 10

y < b

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3 years ago
Find the probability that the person is frequently or occasionally involved in charity work.
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Given the table below which shows the result of a survey that asked 2,881 people whether they are involved in any type of charity work.

\begin{tabular}&#10;{|c|c|c|c|c|c|}&#10; &Frequently&Occassionally&Not at all&Total\\[1ex]&#10;Male&227&454&798&1,479\\&#10;Female &205&450&747&1,402\\&#10;Total&432&904&1,545&2,881&#10;\end{tabular}

Part A:

If a person is chosen at random, the probability that the person is frequently or occassinally involved in charity work is given by

P(being \ frequently \ involved \ or \ being \ occassionally \ involved)\\ \\= \frac{432}{2881} + \frac{904}{2881} = \frac{1336}{2881}=\bold{0.464}



Part B:

If a person is chosen at random, the probability that the person is female or not involved in charity work at all is given by

P(being&#10; \ female \ or \ not \ being \ involved)\\ \\= &#10;\frac{1402}{2881} + \frac{1545}{2881}-\frac{747}{2881} = &#10;\frac{2200}{2881}=\bold{0.764}



Part C:

If a person is chosen at random, the probability that the person is male or frequently involved in charity work is given by

P(being&#10; \ male \ or \ being \ frequently \ involved)\\ \\= &#10;\frac{1479}{2881} + \frac{432}{2881}-\frac{227}{2881} = &#10;\frac{1684}{2881}=\bold{0.585}



Part D:

If a person is chosen at random, the probability that the person is female or not frequently involved in charity work is given by

P(being&#10; \ female \ or \ not \ being \ frequently \ involved)\\ \\= &#10;\frac{1402}{2881} + \frac{904}{2881} + \frac{1545}{2881}-\frac{450}{2881}-\frac{747}{2881} = &#10;\frac{2654}{2881}=\bold{0.921}



Part E:

The events "being female" and "being frequently involved in charity work" are not mutually exclusive because being a female does not prevent a person from being frequently involved in charity work.

Indeed from the table, there are 205 females who are frequently involved in charity work.

Therefore, the answer to the question is "No, because 205 females are frequently involved charity work".
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Answer:

Step-by-step explanation:

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P(Z > 2.5) = .0062

p-value = .0062. The apples are not within the FDA requirement

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3 years ago
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