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2 years ago

How many 12th are in 3 and describe situation

Mathematics

1 answer:

Mashutka [201]2 years ago

6 0

Answer:

1) 36

2) My mom was trying to quiz me and she told me to pick a fraction and a number. I picked 1/12 and 3, she then told me to figure out how many 12ths there are in 3 which I found was 36.

Step-by-step explanation:

3 * 12 = 36

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Any help is appreciated!!

qaws [65]

Answer:

Step-by-step explanation:

600,000,000

6 0

3 years ago

Read 2 more answers

Solve: 3x - 7 - 2x + 5 = 6 A) -8 B) - 18 5 C) 4 D) 8

seropon [69]

Answer:

D) 8

Step-by-step explanation:

3x - 7 - 2x + 5 = 6

3x - 2x = 6 + 7 - 5

x = 8

Note: 6 + 7 -5 = 13 -5 = 8

7 0

3 years ago

Three research departments have 9, 7, and 10 members, respectively. Each department is to select a delegate and an alternate to

aivan3 [116]

Work:
First department: (They have 9 possibilities for delegate and 8 for alternate)
9 * 8 = 72
Second department: (They have 7 possibilities for delegate and 6 for alternate)
7 * 6 = 42
Third department: (They have 10 possibilities for delegate and 9 for alternate)
10 * 9 = 90
Total possibilities:
72 * 42 * 90 = 272,160 possibilities

I hope this helps!

3 0

3 years ago

if an orange gets cut in half how much do you have left if you cut in forths how many slices are there

a_sh-v [17]

Answer:

There are 10 slices left

Step-by-step explanation:

4 0

3 years ago

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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