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2 years ago

8

Algebra 2! Please help ASAP

1 answer:

sweet [91]2 years ago

6 0

the first box = Rational

the answers are the second 2 boxes

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X/2-14=-20 I tried on this one but I failed.

KengaRu [80]

Answer: I’m not going to tell u the answer hrhrhehehehehehrhehsnheaj

Step-by-step explanation:

-34 answer

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3 years ago

Read 2 more answers

How can we predict the sign of the product or quotient of sign numbers?

Andru [333]

Two positives make a positive
two negatives make a positive
1 negative and 1 positive make negative
1 positive and 1 negative make a negative

6 0

2 years ago

Help me with a. b. c. I need help

hammer [34]

Answer:

a) Perimeter = 20.5 Inches

Area is 15 Sq Inches

b) Perimeter is 61.4 Cms

Area is 210 Sq cm

c) Perimeter is 22 yards

Area is 21 Sq Yards

Step-by-step explanation:

a) Perimeter = sum of all the sides

= 8+7.5+5

=20.5

Perimeter = 20.5 Inches

Area of Triangle = $\frac{1}{2} \times height \times Base$

Base = 7.5 In

Height = 4 In

$A = \frac{1}{2} \times 4 \times 7.5$

$A = 2 \times 7.5$

$A = 15$

Hence Area is 15 Sq Inches

b) Perimeter = Sum of all the sides

P=13+23+13.4+12

P=61.4

Perimeter is 61.4 Cms

Area of trapezium = $\frac{1}{2} \times height \times (sum \ of \ parallel \ sides)$

$A= \frac{1}{2} \times 12 \times (12+23)$

$A= 6 \times 25$

$A=210$

Area is 210 Sq cm

c) Perimeter = Sum of all the sides

P = 7+4+7+4 ( as Opposite sides are equal in a parallelogram

P = 22

Perimeter is 22 yards

Area of a parallelogram = $base \times height$

$A= 7 \times 3$

$A = 21$

Area is 21 Sq Yards

3 0

3 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:

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

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

$Pr(5) = 0.267$

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

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,

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

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

$Median = 5$

<em>Person B has a greater median of 5</em>

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

A rectangle has a height of 43 and a width of 3x + 1.

fomenos

A = 43 • 3x + 1
= 44 •3x
= 132x

6 0

3 years ago

Read 2 more answers