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3 years ago

The weight of a stack of standard 8.5x11 copier paper vs number of sheets of paper. Is it proportional

2 answers:

7 0

Answer: The relationship between the weight of a stack of standard 8.5x11 copier paper vs number of sheets of paper is proportional.

Step-by-step explanation:

Let us assume here, in this situation that each page has the same weight so if i want to find the weight of the complete stack, I will multiply the number of sheets of paper by the weight of a single sheet of paper.

galina1969 [7]3 years ago

7 0

Answer:

Yes it is proportional because w=kp k=paper weight which is constant.

Step-by-step explanation:

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PLEASE HELPPPPP THE IS ON A BIG QUIZ The expression below represents the cost, c, of an item that is 25% off c-0.25c

ddd [48]

Answer:

option b is the answer of your question

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2 years ago

Read 2 more answers

mr henderson mixed 3.732g of chemical A, 13.902 of chemical B and 5.208g of Chemical C to make 3 does of medicine, About how muc

Dafna11 [192]

Answer:

22.8 grams

Step-by-step explanation:

To find total amount of medicine, we have to add all the 3 quantities of Chemical A, Chemical B, and Chemical C.

Before we do that, we have to round up each quantity to the nearest tenths.

Nearest tenths means we can take "1 digit" after the decimal point. For that we use rounding rules.

If the 2nd digit to the right of decimal is 5 or greater, we round up the 1st digit to the next whole number. If the 2nd digit is less than 5, we keep the 1st digit same.

So, using that we round up each of the three numbers as shown:

3.732 becomes 3.7 (since 3 is less than 5, we keep the 7)

13.902 becomes 13.9 (since 0 is less than 5, we keep 9)

5.208 becomes 5.2 (since 0 is less than 5, we keep 2)

Now we add:

3.7 + 13.9 + 5.2 = 22.8 grams

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3 years ago

The following 5 questions are based on this information: An economist claims that average weekly food expenditure of households

liq [111]

Answer:

Null hypothesis: $\mu_{1}-\mu_{0}\leq 0$

Alternative hypothesis: $\mu_{1}-\mu_{2}>0$

$SE=\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}=2.705$

b) 2.70

$t=\frac{(164-159)-0}{\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}}=1.850$

b. 1.85

$p_v =P(Z>1.85)=0.032$

b. 0.03

a. We can reject H(0) in favor of H(a)

Step-by-step explanation:

Data given and notation

$\bar X_{1}=164$ represent the mean for the sample 1

$\bar X_{2}=159$ represent the mean for the sample 2

$\sigma_{1}=12.5$ represent the population standard deviation for the sample 1

$s_{2}=9.25$ represent the population standard deviation for the sample B2

$n_{1}=35$ sample size selected 1

$n_{2}=30$ sample size selected 2

$\alpha=0.05$ 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 expenditure of households in City 1 is more than that of households in City 2, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{0}\leq 0$

Alternative hypothesis: $\mu_{1}-\mu_{2}>0$

We know the population deviations, so for this case is better apply a z test to compare means, and the statistic is given by:

$z=\frac{(\bar X_{1}-\bar X_{2})-0}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (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".

Standard error

The standard error on this case is given by:

$SE=\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}$

Replacing the values that we have we got:

$SE=\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}=2.705$

b. 2.70

Calculate the statistic

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

$t=\frac{(164-159)-0}{\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}}=1.850$

P-value

Since is a one side right tailed test the p value would be:

$p_v =P(Z>1.85)=0.032$

b. 0.03

Conclusion

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.