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Janine is packing carrots. Each large box holds 20 2-pound bags of carrots. If janine has 800 bags of carrots, how many boxes can she fill?

Answers

Answer 1

Answer:

Answer: The number of boxes she can fill is 40

Step-by-step explanation:

Given : Janine is packing carrots. Each large box holds 20 2-pound bags of carrots.

i.e. the number of 2-pound bags of carrots in each box = 20

The number of bags of carrot = 800

The number of boxes she can fill is given by :-

$(800)/(20)=40$

Therefore , the number of boxes she can fill is 40.

Answer 2

Answer: This is a division problem:

Each box holds 20 bags.

800 bags / 20 bags per box = # of boxes

40 boxes are needed.

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A true/false test has 90 questions. Suppose a passing grade is 58 or more correct answers. Test the claim that a student knows more than half of the answers and is not just guessing. Assume the student gets 58 answers correct out of 90. Use a significance level of 0.05. Steps 1 and 2 of a hypothesis test procedure are given below. Show step 3, finding the test statistic and the p-value and step 4, interpreting the results.

Answers

Answer:

1 and 2) Null hypothesis: $p \leq 0.5$

Alternative hypothesis: $p > 0.5$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}}$ (1)

3) $z=\frac{0.644 -0.5}{\sqrt{(0.5(1-0.5))/(90)}}=2.732$

4) $p_v =P(z>2.732)=0.0031$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of correct answers is not significantly higher than 0.5

Step-by-step explanation:

Data given and notation

n=90 represent the random sample taken

X=58 represent the number of correct answers

$\hat p=(58)/(90)=0.644$ estimated proportion of correct answers

$p_o=0.5$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

Step 1 and 2: Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion of correct answers is higher than 0.5.:

Null hypothesis: $p \leq 0.5$

Alternative hypothesis: $p > 0.5$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.644 -0.5}{\sqrt{(0.5(1-0.5))/(90)}}=2.732$

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

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

$p_v =P(z>2.732)=0.0031$

Answer:

Step-by-step explanation:

Hello!

The variable of interest is X: the number of correct answers on a true/false test out of 90 questions.

The parameter of interest is p: population proportion of correct answers in a true/false test.

The passing grade is 58/90 correct questions.

The claim is that if the students answer more than half of the answers, then he is not guessing, i.e. if the proportion of correct answers is more than 50%, the student did not guess the answers, symbolically: p>0.5

Then the hypotheses are:

H₀: p ≤ 0.5

H₁: p > 0.5

α: 0.05

since the sample size is large enough, n= 90 questions, you can apply the Central Limit Theorem to approximate the distribution of the sample proportion to normal, p'≈N(p;[p(1-p])/n) and use the standard normal as a statistic:

$Z=\frac{p'-p}{\sqrt{(p(1-p))/(n) } }$ ≈N(0;1)

The sample proportion is the passing grade of the student p': 58/90= 0.64

Then under the null hypothesis the statistic is:

$Z_(H_0)= \frac{0.64-0.5}{\sqrt{(0.5*0.5)/(90) } } = 2.656= 2.66$

This test is one-tailed (right) and so is the p-value, you can calculate it as:

P(Z≥2.66)= 1 - P(Z<2.66)= 1 - 0.996093= 0.003907

With this p-value, the decision is to reject the null hypothesis.

Then at a 5% level, there is significant evidence to conclude that the proportion of correctly answered questions is greater than 50%, this means that the student didn't guess the answers.

I hope this helps!

If a new truck costs $43,750 and it depreciates 18% per year, what will the truck be worth in 5 years? 1 SEE ANSWER

Answers

if it depreciates by 18%, then as each year passes, its worth is reduced to 82% of its value
So, after 5 years, it is worth 43750

The scores on a test are normally distributed. The mean of the test is 750 and the standard deviation is 70. By using the Empirical rule, what scores fall 3 standard deviations from the mean?

a.610 and 820

b.610 and 960

c.680 and 820

d.540 and 960