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3 years ago

A rectangular rug has an area of 31.5 square feet. If the rug is 7 feet wide,

Mathematics

1 answer:

Mandarinka [93]3 years ago

5 0

Answer:

the length of the rug would be 4.5 feet

Step-by-step explanation:

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Evaluate: 32 + (8 − 2) • 4 − 6/3

Ipatiy [6.2K]

The expression to evaluate is 32 + (8 - 2) * 4 - 6/3. In the first step you can solve the parenthesis and the fraction => 32 + (6)*4 - 2. In the second step you mutiply 6*4 => 32 + 24 - 2. Finally, you do the maths: 32 + 24 - 2 = 54. <span>Answer: 54</span>

8 0

3 years ago

I have a friend in need. Her name is Sienna and she has cancer along with co.vid. Please pray for her well being because she rea

velikii [3]

Answer:

She'll be in my prayers

Step-by-step explanation:

When two or three are gathered in His name

3 0

3 years ago

H(t) = -10+ - 6<br>h()= -44

lesya [120]

Answer:

-44

Step-by-step explanation:

4 0

3 years ago

A bag of candy has a mass of 80 g. A chocolate bar has a mass of 110 g. What is the total mass of the bag of candy and the choco

atroni [7]

total mass = mass of a bag of candy + mass of a chocolate bar

total mass = 80g +110g = 190g

3 0

2 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

3 years ago