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Novay_Z [31]
3 years ago
5

The mean number of sick days per employee taken last year by all employees of a large city was 10.6 days. A city administrator i

s investigating whether the mean number of sick days this year is different from the mean number of sick days last year. The administrator takes a random sample of 40 employees and finds the mean of the sample to be 12.9. A hypothesis test will be conducted as part of the investigation.
Which of the following is the correct set of hypotheses?

A. H0:μ=10.6Ha:μ>10.6 AB. H0:μ=10.6Ha:μ≠10.6 BC. H0:μ=10.6Ha:μ<10.6 CD. H0:μ=12.9Ha:μ≠12.9 DE. H0:μ=12.9Ha:μ<12.9 E
Mathematics
1 answer:
Vsevolod [243]3 years ago
7 0

Answer:

H0:μ=10.6

Ha:μ≠10.6

Step-by-step explanation:

you do not find out about the 12.9 until after stating the hypothesis.

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I’ll mark brainliest !!<br> Plz help
saveliy_v [14]

Answer:

3rd option

Step-by-step explanation:

6 0
3 years ago
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Show how the perfect numbers 6 and 28 were generated. Show the aliquot parts of 6 and 28
Mashutka [201]

Step-by-step explanation:

Perfect number is the positive integer which is equal to sum of proper divisors of the number.

Aliquot part is also called as proper divisor which means any divisor of the number which isn't equal to number itself.

<u>Number : 6 </u>

Perfect divisors / Aliquot part = 1, 2, 3

Sum of the divisors = 1 + 2 + 3 = 6

Thus, 6 is a perfect number.

<u>Number : 28</u>

Perfect divisors / Aliquot part = 1, 2, 4, 7, 14

Sum of the divisors = 1 + 2 + 4 + 7 + 14 = 28

Thus, 28 is a perfect number.

3 0
3 years ago
What is the value of -32+(-4+7)(2)?<br> -28<br> -3<br> 3<br> 15
I am Lyosha [343]

Answer:

-26

Step-by-step explanation:

  1. -32+3×2
  2. -32+6
  3. -26

By using BODMAS, any of the answers given is correct!

6 0
2 years ago
Read 2 more answers
Suppose a manufacturer finds that 95% of their production is normal but the final 5% has one or more flaws. Each flawed good has
RUDIKE [14]

Answer:

1)    

FLAW                         TYPE2         NO TYPE2 FLAW

TYPE1                         0.015           0.025

NO TYPE1 FLAW        0.01             0.95

2) 0.04 and $0.04

3) 0.025 and $0.025

4) 0.015 and $0.015

5) 0.95 and $0.95

Step-by-step explanation:

Given that;

financial cost = $1

p(flaw) = 0.05  

p(type 1 flaw / flaw) = 80% = 0.8

p(type 2 flaw / flaw) = 50% = 0.5

p( type 1 and 2 flaw/flaw) = 30% = 0.30

1) Bivariate Table

p( type 1 flaw) = p(flaw) × p(type 1 flaw/flaw) = 0.05 × 0.8 = 0.04

p( type 2 flaw) = p(flaw) × p(type 2 flaw/flaw)  = 0.05 × 0.5 = 0.025

p( type 1 and 2 flaw) =  p(flow) × p( type 1 & 2 flaw/flaw) = 0.05 × 0.3 = 0.015

p( only 1 flow) = 0.04 - 0.015 = 0.025

p( only 2 flow) =  0.025 - 0.015 = 0.01

THEREFORE  the Bivariate Table;

FLAW                         TYPE2         NO TYPE2 FLAW

TYPE1                         0.015           0.025

NO TYPE1 FLAW       0.01              0.95

2) probability and expectations of type 1 flaw?

p( type 1 flaw) = p(flaw) × p(type 1 flaw/flaw) = 0.05 × 0.8 = 0.04

Expected financial cost to the firm per good = $1 × 0.04 = $0.04

3)  probability and expectation of Type 2 flaw

p( type 2 flaw) = p(flaw) × p(type 2 flaw/flaw)  = 0.05 × 0.5 = 0.025

Expected financial cost to the firm per good = $1 × 0.025 = $0.025

4) probability and expectations of Type 1 and 2 flaws

p( type 1 and 2 flaw) =  p(flow) × p( type 1 & 2 flaw/flaw) = 0.05 × 0.3 = 0.015

Expected financial cost to the firm per good = $1 * 0.015 = $0.015

5) probability and expectations of no flaws?

Probability of no flaw = P(No flaw) =95% =  0.95

Expected financial cost saved the firm per good due to no flaw

= $1 × 0.95 = $0.95

5 0
3 years ago
Simplify √5 • √8. <br><br> A.)2√10 <br> B.)4√10 <br> C.)√40 <br> D.)√13
tresset_1 [31]

Answer:

Simplify the radical by breaking the radical  up into a product of known factors.

2 √ 10

Step-by-step explanation:

5 0
3 years ago
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