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makvit [3.9K]
2 years ago
13

Find the limit if it exists

Mathematics
1 answer:
mash [69]2 years ago
3 0

Answer:

20.

Step-by-step explanation:

When x = 2

f(x= 3(2)^3 + 2^2 - 8

= 24 + 4 - 8

= 20.

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If a kitten eats 1/4 cup of food and it’s mom eats 6 times as much food as the kitten how much food did the mom eat?
mariarad [96]

Answer:

The mom ate a cup and a half of food


Step-by-step explanation:


6 0
3 years ago
Explain why the following statement IS NOT a statistical question. "How
deff fn [24]

Answer:

Because it only has one answer and there are no variety of answers.

hope this helps

7 0
2 years ago
How to put 1+0.9+0.08+0.001 in word form
Svet_ta [14]

1.981: One and nine hundred eighty-one thousandths

6 0
2 years ago
In a string of 12 Christmas tree light bulbs, 3 are defective. The bulbs are selected at random and tested, one at a time, until
Juliette [100K]

Using the hypergeometric distribution, it is found that there is a 0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.

In this problem, the bulbs are chosen without replacement, hence the <em>hypergeometric distribution</em> is used to solve this question.

<h3>What is the hypergeometric distribution formula?</h3>

The formula is:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • N is the size of the population.
  • n is the size of the sample.
  • k is the total number of desired outcomes.

In this problem:

  • There are 12 bulbs, hence N = 12.
  • 3 are defective, hence k = 3.

The third defective bulb is the fifth bulb if:

  • Two of the first 4 bulbs are defective, which is P(X = 2) when n = 4.
  • The fifth is defective, with probability of 1/8, as of the eight remaining bulbs, one will be defective.

Hence:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

P(X = 2) = h(2,12,4,3) = \frac{C_{3,2}C_{9,1}}{C_{12,4}} = 0.2182

0.2182 x 1/8 = 0.0273.

0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.

More can be learned about the hypergeometric distribution at brainly.com/question/24826394

8 0
2 years ago
A machine that fills beverage cans is supposed to put 12 ounces of beverage in each can. The standard deviation of the amount in
erica [24]

Answer:

\chi^2 =\frac{10-1}{0.0144} 0.0217 =13.54

The degrees of freedom are:

df=n-1=10-1=9

Now we can calculate the p value using the alternative hypothesis:

p_v =2*P(\chi^2 >13.54)=0.279

Since the p value is higher than the signficance level assumed of 0.05 we have enough evidence to FAIL to reject the null hypothesis and there is no evidence to conclude that the true deviation differs from 0.12 ounces

Step-by-step explanation:

Assuming the following data:"12.14 12.05 12.27 11.89 12.06

12.14 12.05 12.38 11.92 12.14"

We can calculate the sample deviation with this formula:

s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

n=10 represent the sample size

\alpha=0.05 represent the confidence level  

s^2 =0.0217 represent the sample variance

\sigma^2_0 =0.12^2= 0.0144 represent the value to verify

Null and alternative hypothesis

We want to determine whether the standard deviation differs from 0.12 ounce, so the system of hypothesis would be:

Null Hypothesis: \sigma^2 = 0.0144

Alternative hypothesis: \sigma^2 \neq 0.0144

The statistic can be calculated like this;

\chi^2 =\frac{n-1}{\sigma^2_0} s^2

\chi^2 =\frac{10-1}{0.0144} 0.0217 =13.54

The degrees of freedom are:

df=n-1=10-1=9

Now we can calculate the p value using the alternative hypothesis:

p_v =2*P(\chi^2 >13.54)=0.279

Since the p value is higher than the signficance level assumed of 0.05 we have enough evidence to FAIL to reject the null hypothesis and there is no evidence to conclude that the true deviation differs from 0.12 ounces

7 0
3 years ago
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