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Mars2501 [29]
3 years ago
12

3x + 5x = 10 Which problem requires the same strategy (combining like terms)?

Mathematics
2 answers:
harina [27]3 years ago
7 0
3x + 5x = 10 Which problem requires the same strategy (combining like terms)?<span>
3x - 2x = 10

</span>
skad [1K]3 years ago
6 0
The answer is D. 3x - 2x = 10
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Each person will eat 2/3 of a pound of Turkey and 8 people will be at party how many pounds of turkey will be needed
Lelechka [254]
Each person 2/3 pounds of turkey

Since you need the amount for 8 people you multiply 8 times 2/3

8/ 2/. =. 16/ = 5 with 1/3
1. Times 3 3


5 with 1/3 pounds of turkey will be needed
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Answer:

is 6.5y + 10

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A. 8.5<br> B. 8.4<br> C. 8.3<br> D. 8.6 <br><br> @mathstudent55
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C. 8.3

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3 years ago
Leah has meters of cloth, which she divides equally among her friends. Each friend gets meters of cloth. She distributed the clo
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4 0
2 years ago
. Last year, at Northern Manufacturing Company, 200 people had colds during the year. One hundred fifty-five people who did no e
Zielflug [23.3K]

Answer:

(a) \frac{1}{5} or 0.2

(b) \frac{1}{10} or 0.1

(c) \frac{3}{10} or 0.3

(d) They are dependent events. See explanation below.

Step-by-step explanation:

Let C = number of employees that had cold = 200

E = number of employees that exercised = 500

(a) P(C) = \dfrac{200}{1000}=\dfrac{1}{5}=0.2

(b) P(C|E)=\dfrac{\text{Number of employees who exercised and had cold}}{\text{Number of employees who exercised}}=\dfrac{50}{500}=\dfrac{1}{10}=0.1

In probability, the notation P(C|E) means the probability of C given that E has occurred. It is given by

P(C|E)=\dfrac{P(C\cap E)}{P(E)}=\dfrac{n(C\cap E)}{n(E)}

(c) P(C|\sim E) denotes the probability of those who did not exercise but had cold. Using the same notations as in (b),

P(C|\sim E)=\dfrac{n(C\cap \sim E)}{n(E)}= \dfrac{150}{500}=\dfrac{3}{10}=0.3

(d) When two events, C and E, are independent, then

P(C\cap E)=P(C)\timesP(E) (This is the multiplication law of probability).

It means, in English, the probability of C and E occurring is equal to the product of the probability of C and the probability of E. From the question,

P(C)=0.2, P(E)=0.5 (since half the employees exercised)

P(C\cap E)=0.05 (It is the ratio of the number of those that exercised and had cold to the total number of employees)

It is seen that

P(C\cap E)\ne P(C)\timesP(E) since

0.05 \ne 0.2\times0.5

7 0
2 years ago
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