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2 years ago

I need help im not sure how to do this

Mathematics

1 answer:

sleet_krkn [62]2 years ago

3 0

Answer:

Step-by-step explanation:

x-1 = 0

x = 1

f(x) = 3x⁵+2x⁴+5

f(1) = 3(1)⁵+2(1)⁴+5

= 3+2+5

= 10

I hope this helps

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Step-by-step explanation:

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3 years ago

Graph f(x)=3x+5 and g(x)=3x-2

IgorLugansk [536]

Answer:

Two parallel lines

Step-by-step explanation:

the slopes are the same: slope = 3

f(x) 3x + 5 g(x) = 3x - 2

Pick at least three x values and find their corresponding values for y. Remember the f(x) and g(x) represent y

f(x) = 3x + 5 g(x) = 3x - 2

y = 3x + 5 y = 3x - 2

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Locate the ordered pairs and draw the graph on the lines.

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2 years ago

A group of 24 people need transportation from a hotel to a concert. They will be picked up by Uber drivers that can fit 4 passen

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Answer:

Step-by-step explanation:

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2 years ago

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lesya692 [45]

Conditional probability is a measure of the probability of an event given that another event has occurred. If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A|B), or sometimes $P_B(A)$ .

The conditional probability of event A happening, given that event B has happened, written as P(A|B) is given by
$P(A|B)= \frac{P(A \cap B)}{P(B)}$

In the question, we were told that there are three randomly selected coins which can be a nickel, a dime or a quarter.

The probability of selecting one coin is $\frac{1}{3}$

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To find the probability that all three coins are quarters if the first two envelopes Jeanne opens each contain a quarter, let the event that all three coins are quarters be A and the event that the first two envelopes Jeanne opens each contain a quarter be B.

P(A) means that the first envelope contains a quarter AND the second envelope contains a quarter AND the third envelope contains a quarter.

Thus $P(A)= \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27}$

P(B) means that the first envelope contains a quarter AND the second envelope contains a quarter

Thus $P(B)= \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$

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Part B:
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Therefore, $P(C|D)=\left( \frac{ \frac{2}{9} }{ \frac{1}{3} } \right)= \frac{2}{3}$

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