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Doss [256]
2 years ago
12

Anybody know the answer to this problem be correct

Mathematics
1 answer:
REY [17]2 years ago
8 0

Answer:

28 out of 40

2(20 - x)/40 for x questions wrong

Step-by-step explanation:

total points: 40

total Qs: 20

20-6 = 14

14/20 = 28/40

28

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Somebody help pls Complete the table by filling in the multiplicative inverse of the numbers below ​
Harlamova29_29 [7]

Answer:

2---> 1/2

1/5 ----> 5

-4 ----> 1/-4

1 2/5 = 7/5 ----> 1/(7/5) = 5/7

Step-by-step explanation:

8 0
3 years ago
Based on the graphs of f (x) and g(x), what must the domain of (f ⋅ g)(x)be?
dimulka [17.4K]

Answer:

{x ∈ ℝ | x ≠ 2}

Step-by-step explanation:

4 0
2 years ago
PLEASE HELP n+5(-n+1)=7
yKpoI14uk [10]

Answer: n= -1/2

Step-by-step explanation:

4 0
3 years ago
What is the difference of the fractions? Use the number line and equivalent fractions to help find the answer.
ipn [44]

The difference of the fraction is \mathbf{  -\dfrac{3}{4}} by using equivalent expressions.

<h3>What is a number line?</h3>

A number line is a diagrammatic representation of real numbers on the graduated line. It ranges from negative axis to positive axis with 0 being the center position.

A sketch showing the representation of the difference in the mixed fractions can be seen in the image attached below.

Using equivalent expressions to determine the difference between the mixed fractions, we have;

\mathbf{ = -2\frac{1}{2} - (-1\dfrac{3}{4}) }

\mathbf{ = -2\frac{1}{2} +1\dfrac{3}{4} }

\mathbf{ = -\dfrac{5}{2} +\dfrac{7}{4} }

\mathbf{ = -\dfrac{3}{4}}

Learn more about number line here:

brainly.com/question/25230781

#SPJ1

7 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
2 years ago
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