Ask question

Ask question

All categories

3 years ago

14

F(x) = 4/5 x - 2 Find f(3)Answer is a fraction

2 answers:

FromTheMoon [43]3 years ago

7 0

Answer:

2/5

Step-by-step explanation:

f(x) = 4/5 x - 2

Let x=3

f(3) = 4/5 *3 - 2

=12/5 -2

=12/5 -10/5

= 2/5

VMariaS [17]3 years ago

5 0

The answer would be 2/5 it’s like your minusing 2 from 4 and keeping the 5 the dominator and then u would have 2/5

You might be interested in

X = 2y + 4: line equation: please help me

choli [55]

This is how it would look on a graph.

8 0

3 years ago

Hexagon DEFGHI is translated on the coordinate plane below to create hexagon D'E'F'G'H'I':

dem82 [27]

<span>Hexagon DEFGHI is translated on the coordinate plane below to create hexagon D'E'F'G'H'I'
</span> the translation of hexagon DEFGHI to hexagon D'E'F'G'H'I' is<span> represents by the rule
</span><span> d. (x, y)→(x + 7, y − 7)
I ---> </span>(-8, 4)→(x + 7, y − 7)
I' →(-8 + 7, 4 − 7) = (-1,-3)

5 0

3 years ago

How do I eat pears how do I eat pears how do I eat pears how do I eat pears how do I eat pears how do I eat pears how do I eat p

UkoKoshka [18]

Answer: first open your mouth then hold the pear and put it inn your mouth then chew then it will go down your thought

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Three cards are drawn from a standard deck of 52 cards without replacement. Find the probability that the first card is an ace,

MrRissso [65]

Answer:

$4.82\cdot 10^{-4}$

Step-by-step explanation:

In a deck of cart, we have:

a = 4 (aces)

t = 4 (three)

j = 4 (jacks)

And the total number of cards in the deck is

n = 52

So, the probability of drawing an ace as first cart is:

$p(a)=\frac{a}{n}=\frac{4}{52}=\frac{1}{13}=0.0769$

At the second drawing, the ace is not replaced within the deck. So the number of cards left in the deck is

$n-1=51$

Therefore, the probability of drawing a three at the 2nd draw is

$p(t)=\frac{t}{n-1}=\frac{4}{51}=0.0784$

Then, at the third draw, the previous 2 cards are not replaced, so there are now

$n-2=50$

cards in the deck. So, the probability of drawing a jack is

$p(j)=\frac{j}{n-2}=\frac{4}{50}=0.08$

Therefore, the total probability of drawing an ace, a three and then a jack is:

$p(atj)=p(a)\cdot p(j) \cdot p(t)=0.0769\cdot 0.0784 \cdot 0.08 =4.82\cdot 10^{-4}$

4 0

3 years ago

Because all airline passengers do not show up for their reserved seat, an airline sells 125 tickets for a ﬂight that holds only

bekas [8.4K]

Answer: a) 0.9961, b) 0.9886

Step-by-step explanation:

Since we have given that

Probability that does not show up = 0.10

Probability that show up = 0.90

Here, we use "Binomial distribution":

n = 125 and p = 0.90

Number of passengers that hold in a flight = 120

a) What is the probability that every passenger who shows up can take the ﬂight?

$P(X\leq 120)=\sum_{x=0}^{120}^{125}C_x(0.90)^x(0.10)^{125-x}=0.9961$

(b) What is the probability that the ﬂight departs with empty seats?

$P(X\leq 119)=\sum _{x=0}^{119}^{125}C_x(0.90)^x)(0.10)^{125-x}=0.9886$

Hence, a) 0.9961, b) 0.9886

3 0

3 years ago