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Vadim26 [7]
3 years ago
9

Ruth has 6 chores to do. If each chore does not depend on the others, in how many different orders could she do the chores?

Mathematics
2 answers:
Dafna11 [192]3 years ago
7 0
The answer is above you shoukd mark it brainliest
Mice21 [21]3 years ago
3 0
First chore, 6 to choose from.
Second chore, 5 to choose from.
Third chore, 4 to choose from.
Fourth chore, 3 to choose from.
Fifth chore, 2 to choose from.
Sixth chore, 1 to "choose" from.
The total number of orders possible is the product
6*5*4*3*2*1=6!=720
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I think the answer is d. The slope of AC=Slope of DF.

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3 years ago
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The number of windows on each floor form an arithmetic sequence; 124 windows on the first floor, 116 windows on the second floor
zubka84 [21]

Answer:

Formula is 132 - 8n, and there are 52 windows on 10th floor.

Step-by-step explanation:

difference = 116-124 = -8  ( 8 less windows for every floor you go up)

the first fllor has 124 windows

Formula:

a_{1} = 124

a_{n} = 124 + (-8)(n - 1) = 132 - 8n

10th Floor:

a_{10} = 124 + (-8)(10 - 1) = 124 + (-8)(9) = 124 - 72 = 52  windows

6 0
3 years ago
A basketball backboard set that sold for $79 is discounted 15%. what is the sale price​
Georgia [21]

Answer:

67.15

Step-by-step explanation:

Procedure:

The rate is usually given as a percent.

To find the discount, multiply the rate by the original price.

To find the sale price, subtract the discount from original price.

8 0
3 years ago
Bad gums may mean a bad heart. Researchers discovered that 80% of people who have su ered a heart attack had periodontal disease
saul85 [17]

Answer:

0.069 = 6.9% probability that he or she will have a heart attack

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Does not have periodontal disease

Event B: Has a heart attack.

Probability of not having a periodontal disease:

100 - 80 = 20% of 10%(had a heart attack).

30% of 100-10 = 90%(did not have a heart attack). So

P(A) = 0.2*0.1 + 0.3*0.9 = 0.29

Intersection of A and B:

Not having the disease, suffering a heart attack, so 20% of 10%.

P(A cap B) = 0.2*0.1 = 0.02

What is the probability that he or she will have a heart attack?

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.02}{0.29} = 0.069

0.069 = 6.9% probability that he or she will have a heart attack

6 0
3 years ago
Someone help me there’s a time limit on it
Westkost [7]

Answer:

P=3

Step-by-step explanation:

8p =24

P= 24/8

P= 3

4 0
3 years ago
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