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2 years ago

8

Which of the following statements is NOT true?

1 answer:

yanalaym [24]2 years ago

5 0

Answer:

the second one

Step-by-step explanation:

it is not 2/3 because it is 1/2

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7/8 = /48 <br> A. 1 <br> B. 6 <br> C. 42 <br> D. 13

Vsevolod [243]

7/8 = x/48

Cross multiply, which is: a/b = c/d ⇒ ad = bc

After cross multiplying, we get:

8x = 7(48)

8x = 336

Divide 8 on both sides

x = 42

Your final answer is B. 42<span>.</span>

6 0

3 years ago

Help me fill in the blanks please help

agasfer [191]

The missing values in the inequality problem are;

A = -9

B = 72

C = 22

<h3>What is the explanation for the above?</h3>

5p - 8 (p + 1) > 16 ......................... (1) Given

5 (-10) - 8 (-10 +1) > 16 ...................... (2) Given

Where P is equal to 10 [Given in the second line as given in equation 2 above; Hence,

Substitute 10 for p

⇒ 5 (-10) - 8(-10+1)> 16

⇒ -50 -8(-9) > 16

⇒ -50 + 72 > 16

Hence

22 > 16 → T

Please note that T is symbolic of "True" that is

it is true that 22 is greater than 16.

Learn more about inequality problems:

brainly.com/question/11613554

#SPJ1

3 0

2 years ago

Will mark YOU brainliest!!

Akimi4 [234]

Answer: 15 miles per gallon

Step-by-step explanation:

135 divided by 9

5 0

2 years ago

Read 2 more answers

Need answer ASAP:

stealth61 [152]

Answer:

Step-by-step explanation:first you multiply 0.76 by 14then multiply 0.68 by 8 then multiply 6.71 by 4. After you get your answer for each one you add them by the cost of each one

7 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:

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

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

$Pr(5) = 0.267$

<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