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V125BC [204]
3 years ago
6

He puts 1/4 of the treasure to aside for himself he gives 1/2 of what is left to his friend bill he shares 2/3 of the remaining

treasure equally between the rest of the crew he buries the remaining 3 bars on a secret island how many bars did the pirates have
Mathematics
1 answer:
Andrej [43]3 years ago
4 0

Answer:

24 bars of treasure

Step-by-step explanation:

You just need to do the opposite in order to find the total value since you divided the treasure up we need to multiply the treasure “back”.

3 buried bars is equal to 1/3 of the remaining treasure. Multiply by 3 to get 3/3 or 100% of buried and crew portion. So we now have 9 bars. Then multiply by 2 to get 100% of what was given to bill, crew and buried. We now have 18 bars. Then we know that we took 1/4 for ourselves so that means 18 bars represents 3/4 of the treasure. So when we look at factors of 18 we find that 3 x 6 = 18 and 4 x 6 = 24 so we conclude that we had 24 bars of treasure before the split.

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Suppose that, in the past, 40% of all adults favored capital punishment. Do we have reason to believe that the proportion of adu
Alex_Xolod [135]

Answer:

The p-value of the test is 0.1469 > 0.05, which means that there is no reason to believe that the proportion of adults favoring capital punishment today has increased, using a 0.05 level of significance.

Step-by-step explanation:

Suppose that, in the past, 40% of all adults favored capital punishment. Test if the proportion has increased:

At the null hypothesis, we test if the proportion is still of 40%, that is:

H_0: p = 0.4

At the alternative hypothesis, we test if the proportion has increased, that is, is greater than 40%, so:

H_1: p > 0.4

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

0.4 is tested at the null hypothesis:

This means that \mu = 0.4, \sigma = \sqrt{0.4*0.6}

Random sample of 15 adults, 8 favor capital punishment.

This means that n = 15, X = \frac{8}{15} = 0.5333

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{0.5333 - 0.4}{\frac{\sqrt{0.4*0.6}}{\sqrt{15}}}

z = 1.05

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion of 0.5333 or more, which is 1 subtracted by the p-value of z = 1.05.

Looking at the z-table, z = 1.05 has a p-value of 0.8531.

1 - 0.8531 = 0.1469.

The p-value of the test is 0.1469 > 0.05, which means that there is no reason to believe that the proportion of adults favoring capital punishment today has increased, using a 0.05 level of significance.

3 0
3 years ago
Find the lcd of 2/3 and 1/3
wariber [46]
Hello!

The least common denominator is like the least common multiple. As you can see, these numbers already have the same denominators, and cannot be divided any further as they are prime. 

Therefore, our answer is 3.

I hope this helps!
4 0
3 years ago
Read 2 more answers
the graph of y = x2 11x 24 is equivalent to the graph of which equation? a.y = (x 8)(x 3) b.y = (x 4)(x 6) c.y = (x 9)(x 2) d.y
Goshia [24]
Y = x^2 + 11x + 24 = (x + 8)(x + 3)
7 0
3 years ago
Read 2 more answers
The Rocky Mountain News (January 24, 1994) indicated that the 20-year mean snowfall in the Denver/Boulder region is 28.76 inches
ycow [4]

Answer:

The p-value of the test is 0.0007 < 0.05, indicating that the the snowfall for the 1993-1994 winters was higher than the previous 20-year average.

Step-by-step explanation:

20-year mean snowfall in the Denver/Boulder region is 28.76 inches. Test if the snowfall for the 1993-1994 winters has higher than the previous 20-year average.

At the null hypothesis, we test if the average was the same, that is, of 28.76 inches. So

H_0: \mu = 28.76

At the alternate hypothesis, we test if the average incresaed, that is, it was higher than 28.76 inches. So

H_1: \mu > 28.76

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

28.76 is tested at the null hypothesis:

This means that \mu = 28.76

Standard deviation of 7.5 inches. However, for the winter of 1993-1994, the average snowfall for a sample of 32 different locations was 33 inches.

This means that \sigma = 7.5, X = 33, n = 32.

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{33 - 28.76}{\frac{7.5}{\sqrt{32}}}

z = 3.2

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 33, which is 1 subtracted by the p-value of z = 3.2. In this question, we consider the standard level \alpha = 0.05.

Looking at the z-table, z = 3.2 has a p-value of 0.9993.

1 - 0.9993 = 0.0007

The p-value of the test is 0.0007 < 0.05, indicating that the the snowfall for the 1993-1994 winters was higher than the previous 20-year average.

5 0
2 years ago
Which formula would be used to find the measure of angle 1?
Stels [109]
What’s angle 1? U need to post a pic
7 0
3 years ago
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