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Travka [436]
2 years ago
8

A minister claims that more than 60% of the adult population attends a religious service at least once a month. The null and alt

ernative hypotheses you'd use to test this claim would be:
a. H0 : μ = .6, Ha: μ > .6
b. H0 : p = .6, Ha: p > .6
c. H0 : p = .6, Ha: p ≠ .6
d. H0 : pˆ = .6, Ha: pˆ > .6
e. H0 : x = .6, Ha: x > .6
Mathematics
1 answer:
LuckyWell [14K]2 years ago
6 0

Answer:

d. H0 : pˆ = .6, Ha: pˆ > .6

Step-by-step explanation:

Given that a minister claims that more than 60% of the adult population attends a religious service at least once a month.

Here the subject of interest about a proportion of adult population who attends a religious service at least once a month.

The claim is  this proportion is greater than 60% and we want to check  whether this claim is true.

Hence hypotheses would be for proportion only and not mean.

Since population proportion is to be checked we use p hat

Correct answer is

d. H0 : pˆ = .6, Ha: pˆ > .6

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The number w and 0.8 are additive inverses.
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Since, the number w and 0.8 are additive inverses.

A number 'a' is said to have an additive inverse '-a' if "a+ (-a)= 0".

Since, 'w' and '0.8' are additive inverses of each other such that w + 0.8 = 0

Therefore, the value of 'w' should be '-0.8' so that -0.8 + 0.8 = 0.

So, the value of 'w' is =0.8

Now, Refer to the attached image which represents the position of 0.8 , w ( that is -0.8) and the sum of 0.8 and w.

Sum of 0.8 and w = 0.8 + w

= 0.8 +(-0.8)

= 0.

3 0
2 years ago
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The angle measurements in the diagram are represented by the following expressions
Klio2033 [76]
8x-10=3x+90
5x=100
X=20
B=3(20)+90=150
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2 years ago
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In 2014, 85 percent of households in the United States had a computer. For a randomly selected sample of 200 households in 2014,
DochEvi [55]

Answer:

The mean of C is 170 households

The standard deviation of C, is approximately 5 households

Step-by-step explanation:

The given parameters are;

The percentage of households in the United States that had a computer in 2014 = 85%

The size of the randomly selected sample in 2014, n = 200

The random variable representing the number of households that had a computer = C

Therefore, we have;

The probability of a household having a computer P = 85/100 = 0.85

Let

Therefore;

The mean (expected) number in the sample, μₓ, = E(x) = n × P is given as follows;

μₓ = 200 × 0.85 = 170

The mean of C = μₓ = 170

The variance, σ² = n × P × (1 - P) = 200 × 0.85 × (1 - 0.85) = 25.5

Therefore;

The standard deviation, σ = √(σ²) = √(25.5) ≈ 5.05

The standard deviation of C, σ ≈ 5 households (we round (down) to the nearest whole number)

7 0
2 years ago
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For commission as a realtor, Michelle earns $349 plus 3% of the purchase price for each home she helps buy or sell. If she earne
Papessa [141]

Well, let us solve this step by step.

We know that Michelle earns 349 plus 3% of the Purchase price. Let us call the Purchase price as P, so that:

 

Earnings, E = 349 + 0.03 P

 

So if she earns 8,965 (E = 8,965) so we can find P:

 

8,965 = 349 + 0.03 P

0.03 P = 8,616

P = $287,200

8 0
2 years ago
Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar. David invested $220
S_A_V [24]

Answer:

The answer is 23 years.

Step-by-step explanation:

We will use the formula :

A=P(1+r)^{t}

Here P = 220

r = 3%

A = 400

Putting these values in the formula we get,

400=200(1+0.03)^{t}

2=1.03^{t}

Taking log on both sides,

ln(1.03)t=ln 2

t=\frac{ln(2)}{ln(1.03)}

t=23.44 or rounding to nearest, t=23 years

The graph of the function can be shown as below.

3 0
2 years ago
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