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RUDIKE [14]
2 years ago
12

Evaluate a^2b^2? when a = 5 and b = 3.

Mathematics
1 answer:
kvasek [131]2 years ago
5 0

Answer:

D) 7

Step-by-step explanation:

a^{2} -2b^{2}

a=5

b=3

\left(5\right)^2-2\left(3\right)^2

Calculate 5 to the power of 2 and get 25:-

25-2\times 3^{2}

Calculate 3 to the power of 2 to get 9 and then multiply 2 and 9 to get 18:-

25-18

=7

<u>OAmalOHopeO</u>

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Solve the expression 5x+2 1/2=15
Jobisdone [24]
X= 2.5 that is the correct answer
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3 years ago
Pls do it for me thank you so much
Bond [772]

Answer:

(a) Attached to this response.

(b)

<em>(i)</em> <em>0 cups = 0.275</em>

<em>(ii) 4 cups = 0.125</em>

<em>(iii) 8 cups = 0.025</em>

<em />

<em />

Step-by-step explanation:

(a) The frequency and relative frequency table has been attached to this response.

i. The first column, labelled x, is the number of cups of coffee consumed per day.

ii. The second column, labelled f, is the number of people that consumed x cups of coffee per day. It is found by counting the number of occurrences (i.e frequency) of the numbers of the first column, x, in the given data.

For example, 0 in first column appears 11 times in the given data. Also, 5 in the first column appears 4 times in the given data.

iii. The third column, labelled r, is the relative relative frequency of the number of cups of coffee. It is calculated by dividing the frequency of the cups of coffee by the total number of people that consumed it. As shown on the table, it is calculated by dividing the each of the values on the second column by the sum of the values on that same column (second column).

For example, the relative frequency of 2 cups of coffee is given by:

r = 7  ÷ 40 = 0.175

Also, the relative frequency of 4 cups of coffee is given by;

r = 5 ÷ 40 = 0.125

(b) The probability (Pₓ) that a randomly selected person consumed x cups of coffee is given by;

Pₓ = frequency of x ÷ total frequency

This is also the relative frequency of x

Therefore,

<em>(i) The probability (P₀) that a randomly selected person consumed 0 cups of coffee is given by;</em>

the relative frequency of 0<em> = 0.275</em>

<em></em>

<em>(ii) The probability (P₄) that a randomly selected person consumed 4 cups of coffee is given by;</em>

the relative frequency of 4<em> = 0.125</em>

<em>(iii) The probability (P₈) that a randomly selected person consumed 8 cups of coffee is given by;</em>

the relative frequency of 8<em> = 0.025</em>

<em></em>

<em></em>

<em></em>

Download docx
5 0
3 years ago
The probability that an American CEO can transact business in a foreign language is .20. Twelve American CEOs are chosen at rand
NNADVOKAT [17]

Answer with Step-by-step explanation:

We are given that

The probability that an American CEO can transact business in  foreign language=0.20

The probability than an American CEO can not transact business in foreign language=1-0.20=0.80

Total number of American CEOs  chosen=12

a. The probability that none can transact business in a foreign language=12C_0(0.20)^0(0.80)^{12}

Using binomial theorem nC_r(1-p)^{n-r}p^r

The probability that none can transact business in a foreign language=\frac{12!}{0!(12-0)!}(0.8)^{12}=(0.8)^{12}

b.The probability that at least two can transact business in a foreign language=1-P(x=0)-p(x=1)=1-((0.8)^{12}+12C_1(0.8)^{11}(0.2))=1-((0.8)^{12}+12(0.8)^{11}}(0.2))

c.The probability that all 12 can transact business in a foreign language=12C_{12}(0.8)^0(0.2)^{12}

The probability that all 12 can transact business in a foreign language=\frac{12!}{12!}(0.2)^{12}=(0.2)^{12}

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1. (16,25)-(24,5) the interval it falls from is 25 mph to 5 mph or it decreased by 20 mph.

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7 0
3 years ago
An electronic product contains 48 integrated circuits. The probability that any integrated circuit is defective is 0.01, and the
BigorU [14]

Answer:

0.6173 = 61.73% probability that the product operates.

Step-by-step explanation:

For each integrated circuit, there are only two possible outcomes. Either they are defective, or they are not. The integrated circuits are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

An electronic product contains 48 integrated circuits.

This means that n = 48

The probability that any integrated circuit is defective is 0.01.

This means that p = 0.01

The product operates only if there are no defective integrated circuits. What is the probability that the product operates?

This is P(X = 0). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{48,0}.(0.01)^{0}.(0.99)^{48} = 0.6173

0.6173 = 61.73% probability that the product operates.

5 0
3 years ago
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