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brilliants [131]

3 years ago

14

Can someone help me with this??

1 answer:

dusya [7]3 years ago

4 0

Answer:

Carlos and Pamela drove 120 miles on the first day, 240 miles on the second day, and 290 miles on the third day.

Step-by-step explanation:

Let x be the number of miles driven on the first day.

Then they drove twice as many miles, or 2x on the second day

and 50 miles more than the second day's, so 2x + 50

The total is 650 across all three days, so we'll take the sum.

x + 2x + (2x + 50) = 650

Combine like terms on the left

5x + 50 = 650

Subtract 50 on both sides

5x = 600

Divide by 5 on both sides

x = 120

Check work:

120 + 2(120) + 2(120) + 50 = 650

120 + 240 + 240 + 50 = 650

600 + 50 = 650

650 = 650

So they drove 120 miles on the first day

2*120 = 240 miles on the second day

and 2*120 + 50 = 240 + 50 = 290 miles on the third day

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Alex73 [517]

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Step-by-step explanation:

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3 years ago

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Mazyrski [523]

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3 years ago

Someone help me answer this

dolphi86 [110]

Answer:

c = 53 yd is the answer.

Step-by-step explanation:

a = 45 yd

b = 28 yd

c = ?

According to the Pythagorean Theorem,

a² + b² = c²

45² + 28² = c²

2025 + 784 = c²

c² = 2809

c = 53 yd

∴ The dog runs 53 yd

6 0

3 years ago

Find two consecutive numbers with a product of 121

Crank

One number: x
Its consecutive: x + 1

Product:

x(x+1)=121
x² + x - 121 = 0

Δ = 1² - 4.1.(-121)
Δ = 1+484
Δ = 485

As square root of 485 is not integer, do not exist two consecutive numbers with product of 121

6 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