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2 years ago

Jasmine left me so I have to improvise anyways PLS HELP WILL GIVE BRAINLIEST

Mathematics

1 answer:

Nostrana [21]2 years ago

4 0

3 and 1/6

to check your work, if you do 2+1 you get 3. and to do 3 and 1/6 you must do 6 times 3 which is 18 and add 1 bc that is the numerator. so you get 19/6. 6 can go into 19 3 times giving you 18 which will then leave you with 1 making the answer to this question 3 and 1/6. 3 being for how many times 6 can go into 19. 1 being what is left of 19 after 6 fitting into 19 3 times. and 6 being the denominator.

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Please please help me I need help Ill give brainy please help

zzz [600]

Answer:

45 cm

Step-by-step explanation:

$l=\sqrt{r^2+h^2}=\sqrt{27^2+36^2} =45 cm$

7 0

2 years ago

Need help to solve please??<br> Need to proof

inysia [295]

Answer:

Step-by-step explanation:

# Statement Reasoning

1). KJ║ML Given

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3). KL ≅ LK Reflexive property

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8 0

2 years ago

Help plzzzzzzzzzzzzz

ZanzabumX [31]

1 1/2 * 3/4 =

Change 1 1/2 to an improper fraction = 3/2

3/2 * 3/4 = 6/4 * 3/4

6/4 * 3/4 = 18/4 = 4 2/4 = 4 1/2

The answer is 4 1/2.

Hope this helped☺☺

4 0

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

Given that X ~ N(120, 35). We survey samples of 25 and are interested in the distribution of X-bar. Find the z-score associated

lubasha [3.4K]

Answer:

Option c -2.8571

Step-by-step explanation:

z-score are calculated as

$z=\frac{xbar-mean}{\frac{S.D}{\sqrt{n} } }$ .

We are given that mean=120 and standard deviation=S.D=35 as X~N(120,35).

Sample size=n=25.

We have to find z-score for x-bar=100. So,

z-score=[100-120]/[35/√25]

z-score=[-20]/[35/5]

z-score=-20/7

z-score=-2.8571.

Thus, the z-score associated with x-bar = 100 is -2.8571.

8 0

3 years ago