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3 years ago

Hi everybody.Who could help with my gw please:))But when u would write the answets please write thev

Mathematics

1 answer:

podryga [215]3 years ago

3 0

% error = {|Approximate Value - Exact Value| ÷ |Exact Value|} * 100%

The difference of the Approximate and the Exact Value is an absolute value. Therefore, the negative sign of the difference is disregarded.

1) Estimate = 250 g ; Actual = 240 g
(250 - 240)/240 * 100% = 10/240 * 100% = 4.17%

2) Estimate = 97.5°C ; Actual = 100°C
(97.5 - 100)/100 * 100% = |-2.5|/100 * 100/% = 2.5%

3) Estimate = 10.4 ml ; Actual = 9.7 ml
(10.4 - 9.7)/9.7 * 100% = 0.7/9.7 * 100% = 7.22%

4) Estimate = 0.156 cal/g°C ; Actual = 0.185 cal/g°C
(0.156-0.185)/0.185 * 100% = |-0.029|/0.185 * 100% = 15.68%

5) Estimate = 62.5 g ; Actual = 65.2 g
(62.50 - 65.20)/65.20 * 100% = |-2.7|/65.20 * 100% = 4.14%

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2 years ago

Read 2 more answers

What is 14/15 over 4/7 equal to

Aneli [31]

Answer:

$\frac{0.93}{0.57}$

Step-by-step explanation:

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2 years ago

A researcher claims that the mean of the salaries of elementary school teachers is greater than the mean of the salaries of seco

Brut [27]

Answer:

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

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

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

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

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:

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

Conclusion

5 0

2 years ago