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2 years ago

Which do you pick and why? How much money do you end up with in each case?

Mathematics

1 answer:

Mademuasel [1]2 years ago

7 0

Answer:

1. Start with $1 and then double the money you have everyday for 30 days. You would end up with 1,073,741,824 on the 30th day.

Step-by-step explanation:

Why you should choose the first option:

1 x 2 = 2

2 x 2 = 4

4 x 2 = 8

8 x 2 = 16

16 x 2 = 32

32 x 2 = 64

64 x 2 = 128

128 x 2 = 256

256 x 2 = 512

512 x 2 = 1024

1024 x 2 = 2048

2048 x 2 = 4096

4096 x 2 = 8192

8192 x 2 = 16384

16384 x 2 = 32768

32768 x 2 = 65536

65536 x 2 = 131072

131072 x 2 = 262144

262144 x 2 = 524288

524288 x 2 = 1048576

1048576 x 2 = 2097152

2097152 x 2 = 4194304

4194304 x 2 = 8388608

8388608 x 2 = 16777216

16777216 x 2 = 33554432

33554432 x 2 = 67108864

67108864 x 2 = 134217728

134217728 x 2 = 268435456

268435456 x 2 = 536870912

536870912 x 2 = 1073741824

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Answer:

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If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

We can say that at 5% of significance the true mean is not significantly less than 30 so then the claim that true average fuel efficiency is (at least) 30 mpg makes sense

Step-by-step explanation:

Data given and notation

The mean and sample deviation can be calculated from the following formulas:

$\bar X =\frac{\sum_{i=1}^n x_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)}{n-1}}$

$\bar X=29.35$ represent the sample mean

$s=1.365$ represent the sample standard deviation

$n=6$ sample size

$\mu_o =30$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is lower than 30 or no, the system of hypothesis are :

Null hypothesis: $\mu \geq 30$

Alternative hypothesis: $\mu < 30$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{29.35-30}{\frac{1.365}{\sqrt{6}}}=-1.17$

P-value

We need to calculate the degrees of freedom first given by:

$df=n-1=6-1=5$

Since is a one-side left tailed test the p value would given by:

$p_v =P(t_{(5)}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

We can say that at 5% of significance the true mean is not significantly less than 30 so then the claim that true average fuel efficiency is (at least) 30 mpg makes sense

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