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irga5000 [103]
2 years ago
6

Select the option that is a better deal. A Option A: $4 for 16 ounces of turkey Option B: y = 0.22x, where y is the cost and x i

s the number of ounces of turkey
Option A: $75 for 5 CDs

Option B:

First quadrant of a coordinate plane with X axis labeled Number of C D’s and Y axis labeled Cost in dollars. The line Y equals twelve X is graphed.

Option A: y = 1.75x, where y is the cost and
x is the number of pounds of turkey

Option B:

First quadrant of a coordinate plane with X axis labeled Pounds of Turkey and Y axis labeled Cost in dollars. The line Y equals 2 X is graphed.

Option B: $99 for 12 shirts

Option A:

Table of Shirts and Costs in dollars. 2 shirts cost seventeen dollars. 3 shirts cost twenty-five dollars and fifty cents. 5 shirts cost forty-two dollars and fifty cents. 8 shirts cost sixty-eight dollars.
Mathematics
2 answers:
timurjin [86]2 years ago
7 0

Answer:

b

Step-by-step explanation:

grandymaker [24]2 years ago
3 0

A i think, uh im here

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Suppose a system has two modules, A and B, that function independently. Module A fails with probability 0.24 and Module B fails
DiKsa [7]

Answer:

Idk idk idk idk idk idk idk idk

4 0
2 years ago
A sample with a sample proportion of 0.4 and which of the following sizes will produce the widest 95% confidence interval when e
Fed [463]

Answer:

C. 50

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

The higher the margin of error, the wider an interval is.

As the sample size increases, the margin of error decreases. If we want a widest possible interval, we should select the smallest possible confidence interval.

So the correct answer is:

C. 50

6 0
2 years ago
An investor believes that investing in domestic and international stocks will give a difference in the mean rate of return. They
arlik [135]

Answer:

So on this case the 90% confidence interval would be given by -4.137 \leq \mu_1 -\mu_2 \leq 2.087  

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\bar X_1 =2.0233 represent the sample mean 1  

\bar X_2 =3.048 represent the sample mean 2  

n1=15 represent the sample 1 size  

n2=15 represent the sample 2 size  

s_1 =4.893387 population sample deviation for sample 1  

s_2 =5.12399 population sample deviation for sample 2  

\mu_1 -\mu_2 parameter of interest at 0.1 of significance so the confidence would be 0.9 or 90%

We want to test:

H0: \mu_1 = \mu_2

H1: \mu_1 \neq \mu_2

And we can do this using the confidence interval for the difference of means.

Solution to the problem  

The confidence interval for the difference of means is given by the following formula:  

(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}} (1)  

The point of estimate for \mu_1 -\mu_2 is just given by:  

\bar X_1 -\bar X_2 =2.0233-3.048=-1.0247

The degrees of freedom are given by:

df = n_1 +n_2 -2 = 15+15-2=28  

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,28)".And we see that t_{\alpha/2}=\pm 1.701  

Now we have everything in order to replace into formula (1):  

-1.0247-1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=-4.137  

-1.0247+1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=2.087  

So on this case the 90% confidence interval would be given by -4.137 \leq \mu_1 -\mu_2 \leq 2.087  

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

4 0
1 year ago
One week, Abigail earned $396.80 at her job when she worked for 16 hours. If she is paid the same hourly wage, how many hours wo
Delicious77 [7]
$396.80 divided by 16 to find the hourly wage you get $24.80 per hours so she works 1 hour to get that much so then in order to compute the amount of hours she’ll have to work you divide $148.80 by $24.80 to get 6. She has to work 6 hours to earn $148.80.
3 0
2 years ago
Because of the rainy season, the depth in a pond increases 3% each week. Before the rainy season started, the pond was 10 feet d
kari74 [83]
After first week pond will be:
10 + 10* 0.03 = 10*(1.03) feet deep
after second week it will be:

10*(1.03) + 10*(1.03)*0.03 = 10*(1.03)^2

from this we can conclude that after x week its will be:
10*(1.03)^x

Note that when second week starts our starting pond is already greater by 3%. every week we add 3% of prevous weeks size of pond

for x = 8 we get
f(x) = 10*(1.03)^8 = 12.66 or rounded 13 feet

Answer is D.

5 0
2 years ago
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