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3 years ago

8

Find the indicated measure. Round answers to the nearest hundredth.

1 answer:

Ronch [10]3 years ago

7 0

Bro what I Will do is search up you questions on YouTub it will help you a lot

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Marisa bought four bottles of water.Each bottle of water was 95 cents. Write an equation with the same product as the total cost

meriva

Answer:

$5\times 0.76= 3.8$

Step-by-step explanation:

Given: Marisa bought four bottles

Price of each bottle is $0.95.

Solving with the factors as given to find the total cost paid by Marisa.

Total cost= $4\ bottles\times \$ 0.95= \$ 3.8$

∴ Total cost paid by Marisa is $3.8.

As given to write an equation with the same product as the total cost but different factors.

∴ Lets assume the number of bottles bought by Marisa be "5" and cost of each bottle be "x"

We already find the total cost she paid is $3.8.

Now, writing an equation with different factors, as total cost remain same.

$5\times x= 3.8$

Dividing both side by 5

we get x= $ 0.76

∴ The required equation is $5\times 0.76= 3.8$

6 0

3 years ago

Factor the linear expression 16p- 24 using the greatest common factor of

horsena [70]

The Answer should be A

8 0

4 years ago

Read 2 more answers

PLZ HELP ME LLZ I WILL DO ANYTHING

____ [38]

Answer:

X=45

Step-by-step explanation:

45+3x=180

180-45=135

135/3=45

x has to be 45

x=45

4 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

3 years ago

Jayla recently accepted a new job earning $15.50 per hour. Last month, Jayla earned $2294 after four weeks of work. How many hou

V125BC [204]

Assuming the pay last month was from Jayla's new job, it represents pay for weekly hours of ...

$\dfrac{\$2294}{4\,week}\times\dfrac{1\,hours}{\$15.50}=37\,\dfrac{hours}{week}$

Jayla worked an average of 37 hours each week.

4 0

4 years ago