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3 years ago

Expand and simplify 9w(3x-4y)-5w(x+y)

Mathematics

2 answers:

Vesna [10]3 years ago

5 0

$9w(3x-4y)-5w(x+y)=\\
27wx-36wy-5wx-5wy=\\
22wx-41wy$

Llana [10]3 years ago

4 0

$\textbf{The~Answer~~of~~Question~} \\ \\ 
9w(3x-4y)-5w(x+y) \\ \\ 27wx-36wy-5wx-5wy \\ \\ \boxed{22wx-41wy} \\ \\ \textbf{Sancak1907}$

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alexandr402 [8]

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3 years ago

Please answer asap.......... ...

skelet666 [1.2K]

Answer:

Option B is tge right answer.

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3 years ago

Read 2 more answers

This is rlly easy but Im to lazy to do it- I mark brainliest.

kkurt [141]

Answers:

A. 6 Large taxis = 42 seats 9 Small taxis = 36 seats = 78 seats in total

B. 6 Large taxis = $498 + 9 Small taxis = $450 498+450= $948

C. 5 Large taxis and 10 Small taxis

Step-by-step explanation:

A. 6 Large taxis = 42 seats 9 Small taxis = 36 seats = 78 seats in total

If I did 8 small taxis the total number of seats would be 74, so I did one small taxi more to make it fair. There would be seats for everyone but 3 seats extra

B. 6 Large taxis = $498 + 9 Small taxis = $450 498+450=948

C. 5 Large taxis and 10 Small taxis

While the more small taxis there are, the more cheaper it is for Max but the less seats there would be for 75 people, So I did 1 more small taxi and 1 less large taxi.

The total number of seats now is 75 seats which is perfect amount for 75 people

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3 years ago

he amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and s

sp2606 [1]

Complete question:

He amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is

a) less than 8 minutes

b) between 8 and 9 minutes

c) less than 7.5 minutes

Answer:

a) 0.0708

b) 0.9291

c) 0.0000

Step-by-step explanation:

Given:

n = 47

u = 8.3 mins

s.d = 1.4 mins

a) Less than 8 minutes:

$P(X$

P(X' < 8) = P(Z< - 1.47)

Using the normal distribution table:

NORMSDIST(-1.47)

= 0.0708

b) between 8 and 9 minutes:

P(8< X' <9) = $[\frac{8-8.3}{1.4/ \sqrt{47}}< \frac{X'-u}{s.d/ \sqrt{n}} < \frac{9-8.3}{1.4/ \sqrt{47}}]$

= P(-1.47 <Z< 6.366)

= P( Z< 6.366) - P(Z< -1.47)

Using normal distribution table,

$NORMSDIST(6.366)-NORMSDIST(-1.47)$

0.9999 - 0.0708

= 0.9291

c) Less than 7.5 minutes:

P(X'<7.5) = $P [Z< \frac{7.5-8.3}{1.4/ \sqrt{47}}]$

P(X' < 7.5) = P(Z< -3.92)

NORMSDIST (-3.92)

= 0.0000

3 0

2 years ago

In a carnival game, a person wagers $2 on the roll of two dice. if the total of the two dice is 2, 3, 4, 5, or 6 then the pe

Olin [163]

Answer: a loss of 4 cents

Step-by-step explanation:

The probability of rolling a sum of 2, 3, 4, 5, or 6 is $\dfrac{15}{36}$ which earns $2.00

The probability of rolling a sum of 28, 9, 10, 11, or 12 is $\dfrac{15}{36}$ which loses $2.00

The probability of rolling a sum of 7 is $\dfrac{6}{36}$ which loses $0.25

$\bigg(\dfrac{15}{36}\times \$2.00\bigg)+\bigg(\dfrac{15}{36}\times -\$2.00\bigg)+\bigg(\dfrac{6}{36}\times -\$0.25\bigg)=\boxed{-\$0.04}$

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3 years ago