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3 years ago

An iceberg weighs 450,000 lb. If it loses 0.2% of its weight during the day, what is its new weight at the end of the day?

Mathematics

2 answers:

JulsSmile [24]3 years ago

6 0

Answer: The weight of the iceberg at the end of the day will be 449,100 lbs.

Step-by-step explanation: The question is solved by first calculating what the loss of weight is during the day. The question indicates that it will lose .2%. In order to calculate the loss you need to multiply 450,000 x .002, which equals 900. Next, we subtract the weight loss from the original weight to determine the end of day weight. 450,000 - 900 = 449,100 lbs.

fenix001 [56]3 years ago

4 0

Answer: 449,100 lb.

Step-by-step explanation:

Given: The weight of iceberg = 450,000 lb

The percentage of weight iceberg loses during the day = 0.2%=0.002

Now, the decrease in weight of iceberg at the end of the day = $0.002\times 450000=900\ lb$

Therefore, the new weight of iceberg at the end of the day = $450000-900=449,100\ lb$

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$t=\frac{(48250-45630)}{\sqrt{\frac{3900^2}{26}+\frac{5530^2}{24}}}}=1.921$

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Since is a one sided test the p value would be:

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If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the mean for elementary school teachers is significantly higher than the mean for secondary teachers at 5% of significance

Step-by-step explanation:

Data given and notation

$\bar X_{A}=48250$ represent the mean of elementary teachers

$\bar X_{B}=45630$ represent the mean for secondary teachers

$s_{A}=3900$ represent the sample standard deviation for elementary teacher

$s_{B}=5530$ represent the sample standard deviation for secondary teachers

$n_{A}=26$ sample size selected

$n_{B}=24$ sample size selected

$\alpha=0.05$ 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 mean of the salaries of elementary school teachers is greater than the mean of the salaries of secondary school teachers, the system of hypothesis would be:

Null hypothesis: $\mu_{A}-\mu_{B}\leq 0$

Alternative hypothesis: $\mu_{A}-\mu_{B}>0$

We don't know the population deviations, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{A}-\bar X_{B})}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

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

$t=\frac{(48250-45630)}{\sqrt{\frac{3900^2}{26}+\frac{5530^2}{24}}}}=1.921$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{A}+n_{B}-2=26+24-2=48$

Since is a one sided test the p value would be:

$p_v =P(t_{(48)}>1.921)=0.0303$

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