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2 years ago

Find the value of f(7).

Mathematics

2 answers:

uranmaximum [27]2 years ago

7 0

Answer:

f(7) = - 5

Step-by-step explanation:

Locate x = 7 on the x- axis. Go vertically down to meet the graph, then read the corresponding value of y on the y- axis.

The y- value is - 5

Thus

f(7) = - 5

nataly862011 [7]2 years ago

4 0

Answer: go to 10 the over 7 right im pretty sure

Step-by-step explanation:

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BartSMP [9]

Answer:

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3 years ago

I need help with questions 20 through 25 please. (Attachment is below)

mezya [45]

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3 years ago

Write an equation to match each statement. Then solve. #1) 8 less than a number x is -12. #2) 15 more than a number k is 2. #3)

guajiro [1.7K]

#1) x-8= -12
#2) 15+k= 2
#3) x/-7= -20
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3 years ago

How do I start This?

pav-90 [236]

$\bf \begin{array}{lccclll}
&\stackrel{lbs}{amount}&\stackrel{unit}{cost}&\stackrel{total}{value}\\
&------&------&------\\
Raisins&11&2.30&25.3\\
Peanuts&p&4.50&4.5p\\
------&------&------&------\\
mixture&m&3.29&3.29m
\end{array}$

we know that, the mixture amount is the sum of 11 + p, or 11 + p = m.

we also know that, the total value is also a sum of 25.3 + 4.5p = 3.29m.

$\bf \begin{cases}
11+p=\boxed{m}\\
25.3+4.5p=3.29m\\
----------\\
25.3+4.5p=3.29\left( \boxed{11+p} \right)
\end{cases}
\\\\\\
25.3+4.5p=36.19+3.29p\implies 1.21p=10.89\implies p=\cfrac{10.89}{1.21}
\\\\\\
p=\stackrel{lbs}{9}$

5 0

2 years ago

A random experiment was conducted where a Person A tossed five coins and recorded the number of ""heads"". Person B rolled two d

cestrela7 [59]

Answer:

(10) Person B

(11) Person B

(12) $P(5\ or\ 6) = 60\%$

(13) Person B

Step-by-step explanation:

Given

Person A $\to$ 5 coins (records the outcome of Heads)

Person $\to$ Rolls 2 dice (recorded the larger number)

Person A

First, we list out the sample space of roll of 5 coins (It is too long, so I added it as an attachment)

Next, we list out all number of heads in each roll (sorted)

$Head = \{5,4,4,4,4,4,3,3,3,3,3,3,3,3,3,3,2,2,2,2,2,2,2,2,2,2,1,1,1,1,1,0\}$

$n(Head) = 32$

Person B

First, we list out the sample space of toss of 2 coins (It is too long, so I added it as an attachment)

Next, we list out the highest in each toss (sorted)

$Dice = \{2,2,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6\}$

$n(Dice) = 30$

Question 10: Who is likely to get number 5

From person A list of outcomes, the proportion of 5 is:

$Pr(5) = \frac{n(5)}{n(Head)}$

$Pr(5) = \frac{1}{32}$

$Pr(5) = 0.03125$

From person B list of outcomes, the proportion of 5 is:

$Pr(5) = \frac{n(5)}{n(Dice)}$

$Pr(5) = \frac{8}{30}$

$Pr(5) = 0.267$

From the above calculations:  $0.267 > 0.03125$  Hence, person B is more likely to get 5

Question 11: Person with Higher median

For person A

$Median = \frac{n(Head) + 1}{2}th$

$Median = \frac{32 + 1}{2}th$

$Median = \frac{33}{2}th$

$Median = 16.5th$

This means that the median is the mean of the 16th and the 17th item

So,

$Median = \frac{3+2}{2}$

$Median = \frac{5}{2}$

$Median = 2.5$

For person B

$Median = \frac{n(Dice) + 1}{2}th$

$Median = \frac{30 + 1}{2}th$

$Median = \frac{31}{2}th$

$Median = 15.5th$

This means that the median is the mean of the 15th and the 16th item. So,

$Median = \frac{5+5}{2}$

$Median = \frac{10}{2}$

$Median = 5$

Person B has a greater median of 5

Question 12: Probability that B gets 5 or 6

This is calculated as:

$P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}$

From the sample space of person B, we have:

$n(5\ or\ 6) =n(5) + n(6)$

$n(5\ or\ 6) =8+10$

$n(5\ or\ 6) = 18$

So, we have:

$P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}$

$P(5\ or\ 6) = \frac{18}{30}$

$P(5\ or\ 6) = 0.60$

$P(5\ or\ 6) = 60\%$

Question 13: Person with higher probability of 3 or more

Person A

$n(3\ or\ more) = 16$

So:

$P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Head)}$

$P(3\ or\ more) = \frac{16}{32}$

$P(3\ or\ more) = 0.50$

$P(3\ or\ more) = 50\%$

Person B

$n(3\ or\ more) = 28$

So:

$P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Dice)}$

$P(3\ or\ more) = \frac{28}{30}$

$P(3\ or\ more) = 0.933$

$P(3\ or\ more) = 93.3\%$

By comparison:

$93.3\% > 50\%$

Hence, person B has a higher probability of 3 or more

7 0

2 years ago