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SIZIF [17.4K]
2 years ago
8

Suppose seven pairs of similar-looking boots are thrown together in a pile. What is the minimum number of individual boots that

you must pick to be sure of getting a matched pair? Why?Since there are 7 pairs of boots in the pile, if at most one boot is chosen from each pair, the maximum number of boots chosen would be . It follows that if a minimum of Incorrect: Your answer is incorrect. boots are chosen, at least two must be from the same pair.
Mathematics
1 answer:
VikaD [51]2 years ago
8 0

Answer:

We must pick at least 8 individual boots to be sure of picking at least one matching pair as explained from the pigeon hole principle.

Step-by-step explanation:

From pigeonhole principle, if k is a positive integer and k + 1 or more objects are placed into k boxes, then there is at least one box containing 2 or more objects.

Now, since we have 7 pairs of similar looking boots, thus, number of single boots we have will be;

Number of single boots = 7 x 2 = 14

Now, if we select 7 boots from the 14,then there's a possibility of selecting exactly 1 from each pair. Thus, we will not get a matching pair.

Whereas if we select 8 boots from the 14 single boots, then by the pigeon hole principle, at least 2 of the boots will need to be from the same pair. Hence we can pick at least 8 individual boots to be sure of picking at least one matching pair.

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If Isaac uses a coupon entitling him to a 25% discount off the purchase price before tax, how much will his bill be? Assume that
goblinko [34]

Answer:

hope it helps...

Step-by-step explanation:

25/100 x 79

= 19.75

= 79 - 19.75

= $59.25 (before tax)

7/100 x 59.25

= 414.75/100

= 4.15

= 59.25 + 4.15

= $63.4 (after adding tax)

First we have to take out the percentage of discount and minus it from the total cost . To find the discounted price with tax, we have to take out the percentage of the tax from the discounted amount and add it to the discounted amount to get the total cost.

3 0
2 years ago
Adrian Beltre hit 48 home runs during the 2004 Major League Baseball season, but only hit 19 home runs in the 2005 season. By wh
aliya0001 [1]

Answer:

60.42%

Step-by-step explanation:

Number of home runs hit by Adrian Beltre in 2004 = 48

Number of home runs hit by Adrian Beltre in 2005 = 19

To find:

Percentage decrease in home run production from 2004 to 2005.

Solution:

To find the percentage decrease, first of all we need to find the decrease in the number of home runs and then we will divide with the number of home runs in 2004 and then finally will multiply the result with 100 to get the percentage decrease.

Decrease in the number of home runs = Number of home runs in 2004 - Number of home runs in 2005 = 48 - 19 = 29

\text{Percentage of decrease in Home runs} = \dfrac{\text{Decrease in home runs}}{\text{Number of home runs in 2004}}\times 100\\\Rightarrow \text{Percentage of decrease in Home runs} = \dfrac{29}{48}\times 100 \approx \bold{60.42\%}

Therefore, by 60.42% Beltre's home run production has decreased from 2004 to 2005.

4 0
2 years ago
Consider a regenerative vapor power cycle with two feedwater heaters, a closed one and an open one. Steam enters the first turbi
miss Akunina [59]

Answer:

Consider a regenerative vapor power cycle with two feedwater heaters, a closed one and an open one, and reheat. Steam enters the first turbine stage at 12 MPa, 480∘C, and expands to 2 MPa. Some steam is extracted at 2 MPa and fed to the closed feedwater heater. The remainder is reheated at 2 MPa to 440∘C and then expands through the second-stage turbine to 0.3 MPa, where an additional amount is extracted and fed into the open feedwater heater operating at 0.3 MPa. The steam expanding through the third-stage turbine exits at the condenser pressure of 6 kPa. Feedwater leaves the closed heater at 210∘C, 12 MPa, and condensate exiting as saturated liquid at 2 MPa is trapped into the open feedwater heater. Saturated liquid at 0.3 MPa leaves the open feedwater heater. Assume all pumps and turbine stages operate isentropically. Determine for the cycle

a. Draw the cycle on a T-S diagram using the same numbering in the schematic

b. Determine the thermal efficiency of the cycle.

c. Determine the mass flow rate of steam entering the first turbine of the cycle.

(i) Thermal efficiency of the cycle = 43.185 %

(ii) The mass flow rate of steam =93.66 kg/h

Step-by-step explanation:

So we have at

For Point 1 on the T-S diagram we have

p₁ = 80 bar,  

t₁ = 480 °C,

From the super-heated steam tables we have

h₁ = 3349.6 kJ/kg, s₁ = 6.6613 kJ/kg·K

Point 2

p₂ = 20 bar

s₁ = s₂  =with x₂ = (6.6613 -6.6409)/(6.6849-6.6409) = 0.464

therefore h₂ =2953.1 + 0.464×(2977.1 - 2953.1) = 2964.22 kJ/kg

Point 3 on the T-S diagram we have

p₃ = 3 bar again s₁ = s₃  so we go to 3 bar on the steam tables and look up s = 6.6613 kJ/kg·K which is on the saturated steam tables

and x₃ is given as (6.6613 -1.6717)/(6.9916-1.6717) = 0.9379 and

h₃ = 561.43 + x₃×2163.5 = 2590.6 kJ/kg

Point 4

p₄ = 0.08 bar, s₁ = s₄, x₄ = 0.7949 and h₄ = 2083.45 kJ/kg

Point 5  

p₅ = 0.08 bar, h_{f5}= 173.84 kJ/kg

Point 6

Here h₆ is given by  h_{f5} plus the work done to move the water to the open heater therefore h₆ =

= 173.84 kJ/kg + 0.00100848×(3 - 0.08) × 100

= 173.84 kJ/kg + 0.29447616 kJ/kg = 174.13 kJ/kg

Point 7

p₇ = 3 bar, and h_{f7} = 561.43 kJ/kg

Point 8

Here again work is done to convey the fluid t constant pressure thus

h₈ = h_{f7} + v_{f7}× (p₈ - p₇)

561.43 kJ/kg + 0.00107317×(80 - 3)×100 = 569.69 kJ/kg

Point 9

p₉ = 80 bar  and T₉  = 205°C

By interpolating the values on the subcooled teperature tables we get

x₉ = 0.5 and h₉ =  854.94 + 0.5× (899.79 - 854.94) = 877.365 kJ/kg

Point 10

p₁₀ =  20 bar, h₁₀   = h_{f10} = 908.50 kJ/kg

point 11

Here h₁₁ = h₁₀ = 908.50 KJ/kg

For the closed feed water heater, energy and mass flow rate balance gives

m₁ × (h₂ - h₁₀) + (h₈ - h₉) = 0

Therefore m₁ = \frac{ (h_{9}  - h_8)}{(h_{2} - h_{10})}  = 0.14967

while the open water heater we get

m₂×h₃+(1-m₁-m₂)×h₆+m₁×h₁₁ - h₇ = 0

from where m₂ = 0.11479

W_{T} = (h₁-h₂) + (1 - m₁)(h₂ - h₃) +(1 - m₁ - m₂)(h₃ - h₄)

= 1076.11 kJ/kg

W_{p} = (h₈ - h₇) + (1 - m₁ - m₂)×(h₆ - h₅)

= 8.4733 kJ/kg

Q = h₁ -h₉ = 2472.235 kJ/kg

Efficiency = η = \frac{W_{T} - W_{P} }{Q} = 43.185 %

(ii)W_{cycle} = m_1*(W_T -W_P)

m'₁ = 100×10³/1066.63 = 93.66 kg/h

5 0
2 years ago
Use Gauss-Jordan elimination to solve the following linear system: 5x – 2y = –2 –x + 4y = 4 A. (–6,–5) B. (2,0) C. (0,1) D. (–1,
jasenka [17]
Im pretty sure its A.
6 0
2 years ago
Read 2 more answers
Michael arthur deported $2,900 in a new regular savings account that earns 5.5 percent interest compounded semiannually he made
il63 [147K]

Answer:

The amount in the account is $3062.4 (Approx) and compound interest is $162.4 .

Step-by-step explanation:

Formula for compounded semiannually

Amount = P(1 + \frac{r}{2} )^{2t}

Where P is the principle , r is the rate in decimal form and t is the time.

As given

Michael arthur deported $2,900 in a new regular savings account that earns 5.5 percent interest .

Here

Principle =  $2,900

5.5% is written in the decimal form.

= \frac{5.5}{100}

= 0.055

Time = 1 year

Put in the formula

Amount = 2900\times (1 + \frac{0.055}{2} )^{2\times 1}

Amount = 2900\times (1 + 0.0275 )^{2\times 1}

Amount = 2900\times (1.0275)^{2}

Amount = 2900\times 1.056\ (Approx)

Amount = $ 3062.4 (Approx)

Thus the amount after 1 years is $3062.4 (Approx) .

Now find out the compound interest .

Amount = Principle + Compound interest

Here

Amount = $ 3062.4  

Principle = $2,900

Put in the above

$3062.4   = $2,900 + Compound interest

Compound interest = $3062.4 -  $2,900

Compound interest = $162.4






8 0
2 years ago
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