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2 years ago

Betty has the option of using three types of tables in her coffee shop:

Mathematics

1 answer:

Mkey [24]2 years ago

5 0

Answer:

The rectangular table

Step-by-step explanation:

Square table: 36*36=1296

Rectangular table: 20*42=1460 $in^{2}$

Circular table: $\pi * 21^{2}$ ≅1384

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Solve for x: 5x + 15 = 75

yuradex [85]

Answer:

x = 12

Step-by-step explanation:

5x + 15 = 75

-15 -15

5x = 60

x = 12

3 0

3 years ago

Plsss help ASAP I DONT have timeeeee

vitfil [10]

Answer:

It will still be a frame

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Which is the simplified form of the expression 3(7/5x+4) - 2(3/2 - 5/4x)?

Kruka [31]

Answer:

B. $\frac{67}{10} x + 9$

Step-by-step explanation:

Simplify the expression. Remember to go by the order of operations, or PEMDAS.

1) First, distribute the numbers outside of the set of parentheses to the terms inside the set of parentheses next to them. Then, simplify the fractions.

$3(\frac{7}{5} x + 4) -2 (\frac{3}{2} -\frac{5}{4} x)$

$\frac{21}{5}x + 12 - \frac{6}{2} + \frac{10}{4} x$

$\frac{21}{5} x + 12 - 3 + \frac{10}{4} x$

2) Finally, combine like terms. (This means to add or subtract constants and to add or subtract terms with the same variables.) You may need to convert terms to the same denominator in order to do so easier. Then, reduce the fraction.

$\frac{21}{5} x + 9 + \frac{10}{4}x \\\frac{84}{20}x + \frac{50}{20}x + 9 \\\frac{134}{20}x + 9 \\\frac{67}{10}x + 9$

Thus, the answer is $\frac{67}{10} x + 9$ .

8 0

3 years ago

Jamya is making a rectangular blanket. The length of the blanket is 10inches greater than it's worth, W, in inches. Write the fu

natta225 [31]

Answer: (w² + 10)inches²

Step-by-step explanation:

Since the width of the rectangular blanket is given as w and the length of the blanket is 10inches greater than it's width, therefore the length will be:

= 10 + w.

Therefore,

length = 10 + w

width = w

Area = length × width

Area = (10 + w) × w

Area = 10w + w²

Therefore, the area of the blanket will be (w² + 10)inches².

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:

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

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

$Pr(5) = 0.267$

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