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Ivahew [28]
3 years ago
7

The value of 4c + 5g if c = 2 and g = 1.5

Mathematics
1 answer:
bixtya [17]3 years ago
4 0

Answer:

15.5

Step-by-step explanation:

4c + 5g

= 4(2) + 5(1.5)

= 8 + 7.5

= 15.5

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2.7

Step-by-step explanation:

322/7=46

46*125=2.71739

round off

2.7

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Item 19 A plant has an initial height of 1 inch and grows at a constant rate of 3 inches each month. A second plant that also gr
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A line passes through the point (3,9) and has a slope of 4
expeople1 [14]
Point-slope form is y - y1 = m(x - x1). We just need to plug in our stuff. This gives us y - 9 = 4(x - 3). I am not sure if you need to simplify, but if you do, y - 9 = 4x - 12.
7 0
2 years ago
Read 2 more answers
I need help please 36 points
geniusboy [140]

Using the power of zero property, we find that:

a) The simplification of the given expression is 1.

b) Since , equivalent expressions are:  and .

--------------------------------

The power of zero property states that any number that is not zero elevated to zero is 1, that is:

Thus, at item a, , thus the simplification is .

At item b, equivalent expressions are found elevating non-zero numbers to 0, thus  and .

5 0
2 years ago
A student takes an exam containing 1414 multiple choice questions. The probability of choosing a correct answer by knowledgeable
Readme [11.4K]

Answer:

0.0082 = 0.82% probability that he will pass

Step-by-step explanation:

For each question, there are only two possible outcomes. Either the students guesses the correct answer, or he guesses the wrong answer. The probability of guessing the correct answer for a question is independent of other questions. So 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.

In this problem we have that:

n = 14, p = 0.3.

If the student makes knowledgeable guesses, what is the probability that he will pass?

He needs to guess at least 9 answers correctly. So

P(X \geq 9) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14)

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

P(X = 9) = C_{14,9}.(0.3)^{9}.(0.7)^{5} = 0.0066

P(X = 10) = C_{14,10}.(0.3)^{10}.(0.7)^{4} = 0.0014

P(X = 11) = C_{14,11}.(0.3)^{11}.(0.7)^{3} = 0.0002

P(X = 12) = C_{14,12}.(0.3)^{12}.(0.7)^{2} = 0.000024

P(X = 13) = C_{14,13}.(0.3)^{13}.(0.7)^{1} = 0.000002

P(X = 14) = C_{14,14}.(0.3)^{14}.(0.7)^{0} \cong 0

P(X \geq 9) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) = 0.0066 + 0.0014 + 0.0002 + 0.000024 + 0.000002 = 0.0082

0.0082 = 0.82% probability that he will pass

6 0
2 years ago
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