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andrezito [222]
3 years ago
7

Claim: Most adults would erase all of their personal information online if they could. A software firm survey of 563 randomly se

lected adults showed that 60% of them would erase all of their personal information online if they could. Express the original claim in symbolic form. Let the parameter represent the adults that would erase their personal information.Find the value of the test statistic.

Mathematics
1 answer:
sweet-ann [11.9K]3 years ago
5 0

Answer:

a)Claims: P>0.5("MOST" of the adult erase their personal information)

b)Statistics test= 4.746

Step-by-step explanation:

we were given {p}=60% =0.60

where n= 563

since most of adults erase all their personal information online, then (p> 0.5

p=0.5

q=1-p=1-0.5=0.5

CHECK THE ATTACHMENT FOR DETAILED EXPLANATION

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Janet is designing a frame for a client. She wants to prove to her client that m∠AGE ≅ m∠CHE in her sketch. What is the missing
Amiraneli [1.4K]

Answer:

I believe the answer is transative property. I'm currently taking the test so I'm not 100% sure if I'm right yet.

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6 0
3 years ago
Consider a large shipment of 400 refrigerators, of which 40 have defective compressors. If X is the number among 15 randomly sel
Vlad [161]

Answer:

The required probability is 0.94

Step-by-step explanation:

Consider the provided information.

There are 400 refrigerators, of which 40 have defective compressors.

Therefore N = 400 and X = 40

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\frac{40}{400}=0.10

It is given that If X is the number among 15 randomly selected refrigerators that have defective compressors,

That means n=15

Apply the probability density function.

P(X=x)=^nC_xp^x(1-p)^{n-x}

We need to find P(X ≤ 3)

P(X\leq3) =P(X=0)+P(X=1)+P(X=2)+P(X=3)\\P(X\leq3) =\frac{15!}{15!}(0.1)^0(1-0.1)^{15}+\frac{15!}{14!}(0.1)^1(1-0.1)^{14}+\frac{15!}{13!2!}(0.1)^2(1-0.1)^{13}+\frac{15!}{12!3!}(0.1)^3(1-0.1)^{12}\\

P(X\leq3) =0.944444369992\approx 0.94

Hence, the required probability is 0.94

4 0
3 years ago
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