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4 years ago

Answer the following questions CORRECTLY I will know if this is wrong. I WILL REPORT ANY INCORRECT ANSWERS!

Mathematics

2 answers:

Serjik [45]4 years ago

4 0

Answer by JKismyhusbandbae:

|x| - 2 > 0

|x| -2 + 2 > 0 + 2

|x| > 2

Answer: x < - 2 or x > 2

grandymaker [24]4 years ago

3 0

|x| - 2 > 0

first step is to get the absolute value alone

add 2 to each side

|x| - 2 +2> 0+2

|x| > 2

then separate the absolute value in to inequalities, one positive and one negative (the negative flips the inequality)

x >2 or x<-2

Answer: x >2 or x<-2

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3 years ago

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nydimaria [60]

Mark is correct because taking a slice from each pizza would be 3/6 or 1/2 of the pizza but taking two slices of the first pizza, Mark wouldn't get as much because that is the same as 2/6 or 1/3 of the pizza. 1/2 > 1/3 therefore Marks conclusion is correct.

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3 years ago

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 12.3 years, and standard deviation of

Zolol [24]

Answer:

probability of lasting longer = 1.7%

Step-by-step explanation:

We are given:

x' = 14 years

μ = 12.3 years

s = 0.8 years

Thus, let's use the formula for the Z-score value which is;

z = (x' - μ)/s

Thus;

z = (14 - 12.3)/0.8

z = 2.125

From the z-distribution table attached, the p-value is ;

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5 0

3 years ago

The use of social networks has grown dramatically all over the world. In a recent sample of 24 American social network users and

Nady [450]

Answer:

The 99% confidence interval for the true mean amount of time Americans spend social networking each day is (3.02 hours, 3.36 hours).

Step-by-step explanation:

The (1 - α)% confidence interval for population mean when the population standard deviation is not known is:

$CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}$

The information provided is:

$n=24\\\bar x=3.19\ \text{hours}\\s=0.2903\ \text{hours}$

Confidence level = 99%.

Compute the critical value of t for 99% confidence interval and (n - 1) degrees of freedom as follows:

$t_{\alpha/2, (n-1)}=t_{0.01/2, (24-1)}=t_{0.005, 23}=2.807$

*Use a t-table.

Compute the 99% confidence interval for the true mean amount of time Americans spend social networking each day as follows:

$CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}$

$=3.19\pm 2.807\times \frac{0.2903}{\sqrt{24}}\\\\=3.19\pm 0.1663\\\\=(3.0237, 3.3563)\\\\\approx (3.02, 3.36)$

Thus, the 99% confidence interval for the true mean amount of time Americans spend social networking each day is (3.02 hours, 3.36 hours).

6 0

3 years ago