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3 years ago

Long-grain rice costs $0.72/lb. How much would 2.7 lb of long-grain rice cost?

2 answers:

7 0

If $.72 per 1lb, then $x per 2.7lb, by using proportional property:

.72/1 = x/2.7 cross multiply

x = (.72)*(2.7) = $1.94

sweet-ann [11.9K]3 years ago

7 0

Answer:

The cost of the 2.7 lb of long - grain rice is $1.94 .

Option (C) is correct .

Step-by-step explanation:

As given

Long-grain rice costs $0.72/lb.

i.e Cost of long - grain per pound is $0.72 .

Now find out the cost for 2.7 lb long - grain rice .

Cost of 2.7 lb long - grain rice = 2.7 × Cost per pound

Putting values in the above

= 2.7 × 0.72

= $ 1.94 (Approx)

Therefore the cost of the 2.7 lb of long - grain rice is $1.94 .

Option (C) is correct .

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A 45-45-90 triangle has a hypotenuse of length 7. What is the length of one of its legs? If necessary, round your answer to two

harkovskaia [24]

If the legs are legnth x, then the hyptonuse is x√2

x√2=7
divide both sides by √2
x=7/√2
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3 years ago

Find the area and the circumference of the circle. Round your answers to the nearest hundredth.

Alik [6]

Answer:

$area = 200.96 \: {units}^{2} \\ circumfrence = 50.24 \: units$

Step-by-step explanation:

The formula to find the area and circumference of a circle is:

$area = \pi {r}^{2} \\ circumfrence = 2\pi \: r$

*note that pi is 3.14* Therefore the area is:

$formula = \pi {r}^{2} \\ 3.14 \times {8}^{2} = \\ 3.14 \times 64 = 200.96$

Therefore the circumference is:

$formula = \: 2\pi \: r \\ (2)(3.14)(8) = \\ 3.14 \times 16 = 50.24$

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2 years ago

Please simplify –z -10y +2 +15y – 2z -10

Sladkaya [172]

Answer:

25y-3z-8

Step-by-step explanation:

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3 years ago

Read 2 more answers

what is the distance between the points (23,-33) and (4,9) if necessary round your answer to two decimal places

Rufina [12.5K]

The distance between the points (23,-33) and (4,9) rounded to two decimal places is 46.10.

<h3>What is the distance between the points (23,-33) and (4,9)?</h3>

The distance between two points on a graph can be determined using the equation;

D = √[ ( x₂ - x₁ )² + ( y₂ - y₁ )² ]

Given that;

x₁ = 23
x₂ = 4
y₁ = -33
y₂ = 9

We substitute our values into the equation above.

D = √[ ( x₂ - x₁ )² + ( y₂ - y₁ )² ]

D = √[ ( 4 - 23 )² + ( 9 - (-33) )² ]

D = √[ ( -19 )² + ( 42 )² ]

D = √[ 361 + 1764 ]

D = √[ 2125 ]

D = 46.10

Therefore, the distance between the points (23,-33) and (4,9) rounded to two decimal places is 46.10.

Learn more about distance formula here: brainly.com/question/7592016

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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

3 years ago