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Pavlova-9 [17]
3 years ago
13

Describe how you would use the rules of exponents to simplify (7x?yz) 3. You may indicate an exponent in your answer with ^. For

Mathematics
1 answer:
SVETLANKA909090 [29]3 years ago
5 0

Answer:

343x^6y^3z^3

Step-by-step explanation:

One of the rules of exponents demands:

(abc)^m=a^mb^mc^m

where a, b, c and m are coefficients.

Furthermore, you use the following rule for exponents of variables that already have an exponent:

(a^n)^m=a^{m*n}

Thus, you can apply this rules in the following way:

(7x^2yz)^3=(7)^3(x^2)^3(y)^3(z)^3=7*7*7x^{2*3}y^3z^3=343x^6y^3z^3

hence, the answer is 343x^6y^3z^3

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A sample of men's heights was taken and the mean was 68.8 inches. The standard deviation is 2.8 inches. What percent of the men
jarptica [38.1K]

Answer:

87.29%

Step-by-step explanation:

Given: Mean= 68.8 inches

           Standard deviation= 2.8 inches

           

Now, finding the percent of the men in the sample were greater than 72 inches.

We know, z-score= \frac{x-mean}{standard\ deviation}

z-score= \frac{72-68.8}{2.8}

⇒ z-score= \frac{3.2}{2.8} = 1.14

∴ z-score= 1.14

Next, using normal distribution table to find percentage.

Coverting 0.8729 into percentage= 0.8729\times 100

We get the percentage as 87.29%

Hence, 87.29% of the men in the sample were greater than 72 inches.

6 0
3 years ago
All the fourth-graders in a certain elementary school took a standardized test. A total of 85% of the students were found to be
Aneli [31]

Answer:

There is a 2% probability that the student is proficient in neither reading nor mathematics.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a student is proficient in reading

B is the probability that a student is proficient in mathematics.

C is the probability that a student is proficient in neither reading nor mathematics.

We have that:

A = a + (A \cap B)

In which a is the probability that a student is proficient in reading but not mathematics and A \cap B is the probability that a student is proficient in both reading and mathematics.

By the same logic, we have that:

B = b + (A \cap B)

Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So

(A \cup B) + C = 1

In which

(A \cup B) = a + b + (A \cap B)

65% were found to be proficient in both reading and mathematics.

This means that A \cap B = 0.65

78% were found to be proficient in mathematics

This means that B = 0.78

B = b + (A \cap B)

0.78 = b + 0.65

b = 0.13

85% of the students were found to be proficient in reading

This means that A = 0.85

A = a + (A \cap B)

0.85 = a + 0.65

a = 0.20

Proficient in at least one:

(A \cup B) = a + b + (A \cap B) = 0.20 + 0.13 + 0.65 = 0.98

What is the probability that the student is proficient in neither reading nor mathematics?

(A \cup B) + C = 1

C = 1 - (A \cup B) = 1 - 0.98 = 0.02

There is a 2% probability that the student is proficient in neither reading nor mathematics.

6 0
3 years ago
What is 4/7x + 5/14x = 39 ?
Sladkaya [172]

Answer:

yes

Step-by-step explanation:

if you break it down on paper ( which btw i get why you didnt want to do it) you get 39

5 0
3 years ago
Read 2 more answers
Write as a product of two polynomials.<br> a(b–c)+d(c–b)
Naya [18.7K]

Answer:

(b - c)(a - d)

Step-by-step explanation:

Given

a(b - c) + d(c - b) ← factor out - 1

= a(b - c) - d(b - c) ← factor out (b - c) from each term

= (b - c)(a - d)

4 0
3 years ago
Plz stop doing this not cool
Andrew [12]

Answer: So True

Step-by-step explanation:

4 0
2 years ago
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