A hot air balloon holds 74,000 cubic meters of helium, a very noble gas with the density of 0.1785 kilograms per cubic meter. How many kilograms of helium does the balloon contain?

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Answer 1

Answer: P=m/V
Where p is density,m is mass,v is volume
m=PV
m=74000×0.1785
m=13209kg

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2. In an industrial training program, students have been averaging about 64 points on a standardized test. The lecture system was replaced by teaching machines with a lab instructor. There was some doubt as to whether the scores would decrease, increase, or stay the same. A sample of n = 60 students using the teaching machines was tested, resulting in a mean of 68 and a standard deviation of 12. Perform a hypothesis test to see if scores would decrease, increase, or stay the same. Use α = 0.05. Be sure to:1. State your hypotheses.
2. Find the value of the Test Statistic.
3. Find the p-value
4. State your decision (Reject or not)
5. State your conclusion.

Answers

Answer:

Case I

Null hypothesis: $\mu = 64$

Alternative hypothesis: $\mu \neq 64$

$t=(68-64)/((12)/(√(60)))=2.582$

$df=n-1=60-1=59$

Since is a two sided test the p value would given by:

$p_v =2*P(t_((59))>2.582)=0.012$

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 reject the null hypothesis.

We can say that at 5% of significance the true mean is different from 64.

Case II

Null hypothesis: $\mu \leq 64$

Alternative hypothesis: $\mu > 64$

The statistic not changes but the p value does and we have:

$p_v =P(t_((59))>2.582)=0.006$

And we reject the null hypothesis on this case.

So we can conclude that the true mean is significantly higher than 64 at 5% of singnificance

Step-by-step explanation:

Data given and notation

$\bar X=68$ represent the sample mean

$s=12$ represent the sample standard deviation

$n=60$ sample size

$\mu_o =64$ 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 population mean is different from 64 the system of hypothesis are :

Null hypothesis: $\mu = 64$

Alternative hypothesis: $\mu \neq 64$

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:

$t=(\bar X-\mu_o)/((s)/(√(n)))$ (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:

$t=(68-64)/((12)/(√(60)))=2.582$

P-value

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

$df=n-1=60-1=59$

Since is a two sided test the p value would given by:

$p_v =2*P(t_((59))>2.582)=0.012$

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 reject the null hypothesis.

We can say that at 5% of significance the true mean is different from 64.

Now let's assume that we want to see if the mean is significantly higher than 64

Null hypothesis: $\mu \leq 64$

Alternative hypothesis: $\mu > 64$

The statistic not changes but the p value does and we have:

$p_v =P(t_((59))>2.582)=0.006$

And we reject the null hypothesis on this case.

So we can conclude that the true mean is significantly higher than 64 at 5% of singnificance

In need of help asap! ? With steps

Answers

Answer:

<BAE is congruent to <DAC (first choice)

Step-by-step explanation:

Triangle ACD has three angles: <DAC, <ACD, <ADC

Triangle ABE has three angles: <BAE, <AEB, <ABE

Angles ACD, ADC, AEB and ABE are all different angles. Look them up in the figure, and you'll see they are 4 different angles.

Angles BAE and DAC are the same angle.

Answer: <BAE is congruent to <DAC (first choice)

Help i have to turn this in 3 min someone????

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Answer:

Step-by-step explanation:

A manufacturing company regularly conducts quality control checks at specified periods on the products it manufactures. Historically, the failure rate for LED light bulbs that the company manufactures is 5%. Suppose a random sample of 10 LED light bulbs is selected. What is the probability that a) None of the LED light bulbs are defective? b) Exactly one of the LED light bulbs is defective? c) Two or fewer of the LED light bulbs are defective? d) Three or more of the LED light bulbs are not defective?

Answers

Answer:

a) There is a 59.87% probability that none of the LED light bulbs are defective.

b) There is a 31.51% probability that exactly one of the light bulbs is defective.

c) There is a 98.84% probability that two or fewer of the LED light bulbs are defective.

d) There is a 100% probability that three or more of the LED light bulbs are not defective.

Step-by-step explanation:

For each light bulb, there are only two possible outcomes. Either it fails, or it does not. This means that we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)$

In which $C_(n,x)$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_(n,x) = (n!)/(x!(n-x)!)$

And p is the probability of X happening.

In this problem we have that:

$n = 10, p = 0.05$

a) None of the LED light bulbs are defective?

This is P(X = 0).

$P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)$

$P(X = 0) = C_(10,0)*(0.05)^(0)*(0.95)^(10) = 0.5987$

There is a 59.87% probability that none of the LED light bulbs are defective.

b) Exactly one of the LED light bulbs is defective?

This is P(X = 1).

$P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)$

$P(X = 1) = C_(10,1)*(0.05)^(1)*(0.95)^(9) = 0.3151$

There is a 31.51% probability that exactly one of the light bulbs is defective.

c) Two or fewer of the LED light bulbs are defective?

This is

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = 2) = C_(10,2)*(0.05)^(2)*(0.95)^(8) = 0.0746$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5987 + 0.3151 + 0.0746 0.9884$

There is a 98.84% probability that two or fewer of the LED light bulbs are defective.

d) Three or more of the LED light bulbs are not defective?

Now we use p = 0.95.

Either two or fewer are not defective, or three or more are not defective. The sum of these probabilities is decimal 1.

$P(X \leq 2) + P(X \geq 3) = 1$

$P(X \geq 3) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = 0) = C_(10,0)*(0.95)^(0)*(0.05)^(10)\cong 0$

$P(X = 1) = C_(10,1)*(0.95)^(1)*(0.05)^(9) \cong 0$

$P(X = 2) = C_(10,1)*(0.95)^(2)*(0.05)^(8) \cong 0$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0$

$P(X \geq 3) = 1 - P(X \leq 2) = 1$

There is a 100% probability that three or more of the LED light bulbs are not defective.

Final answer:

The question relates to binomial distribution in probability theory. The probabilities calculated include those of none, one, two or less, and three or more LED bulbs being defective out of a random sample of 10.

Explanation:

This question relates to the binomial probability distribution. A binomial distribution is applicable because there are exactly two outcomes in each trial (either the LED bulb is defective or it's not) and the probability of a success remains consistent.

a) In this scenario, 'none of the bulbs being defective' means 10 successes. The formula for probability in a binomial distribution is p(x) = C(n, x) * [p^x] * [(1-p)^(n-x)]. Plugging in the values, we find p(10) = C(10, 10) * [0.95^10] * [0.05^0] = 0.5987 or 59.87%.

b) 'Exactly one of the bulbs being defective' implies 9 successes and 1 failure. Following the same formula, we get p(9) = C(10, 9) * [0.95^9] * [0.05^1] = 0.3151 or 31.51%.

c) 'Two or less bulbs being defective' means 8, 9 or 10 successes. We add the probabilities calculated in (a) and (b) with that of 8 successes to get this probability. Therefore, p(8 or 9 or 10) = p(8) + p(9) + p(10) = 0.95.

d) 'Three or more bulbs are not defective' means anywhere from 3 to 10 successes. As the failure rate is low, it's easier to calculate the case for 0, 1 and 2 successes and subtract it from 1 to find this probability. This gives us p(>=3) = 1 - p(2) - p(1) - p(0) = 0.98.