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Lorico [155]
3 years ago
12

PLEASE HELP!

Mathematics
1 answer:
kotykmax [81]3 years ago
6 0
First off, you can rewrite your equation like so: c=(a-b)x. You can then plug in your given constraints for your choices. If a-b = 1 and c = 0 leaving: 0=1(x), x must equal 0 and only 0 as any constant multiplied by 0 equals 0. So that choice is eliminated. Now let's consider when a=b and c != 0. Since we are given a-b and a=b and c != 0, we have:
c = 0x. This contradicts our claim we made about our constraints. C cannot equal zero but we have a-b=0. Therefore, this claim makes no sense as any value for x will not satisfy the equation. This choice is valid. When a=b and c=0, we have: 0 = 0x. Here, x can be any value and still return 0 as an answer. This choice is valid. If a-b=1 and c != 1, we have: c = 1x. Our only rule here is that c cannot equal 1. This means that x can be any value other than 1 so this choice can be marked down. If a != b and c=0, this gives: 0 = (a-b)x. Given that a-b can be any value, x must be equal to only 0 to satisfy this equation so this choice can't be correct. So the right answers are: option 2, option 3, option 4 and option 5.
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Answer:

Kindly check explanation

Step-by-step explanation:

Given :

Sample size, n = 30

Tcritical value = 2.045

Null hypothesis :

H0: μ = 9.08

Alternative hypothesis :

H1: μ≠ 9.08

Sample mean, m = 8.25

Samole standard deviation, s = 1.67

Test statistic : (m - μ) ÷ s/sqrt(n)

Test statistic : (8.25 - 9.08) ÷ 1.67/sqrt(30)

Test statistic : - 0.83 ÷ 0.3048988

Test statistic : - 2.722

Tstatistic = - 2.722

Decision region :

Reject Null ; if

Tstatistic < Tcritical

Tcritical : - 2.045

-2.722 < - 2.045 ; We reject the Null

Using the α - level (confidence interval) 0.05

The Pvalue for the data from Tstatistic calculator:

df = n - 1 =. 30 - 1 = 29

Pvalue = 0.0108

Reject H0 if :

Pvalue < α

0.0108 < 0.05 ; Hence, we reject the Null

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notsponge [240]
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\\ \sf\longmapsto A=πr^2

\\ \sf\longmapsto A=3.14(5)^2

\\ \sf\longmapsto A=3.14(25)

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