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3 years ago

Evaluate the expression for n = 4. 3(n + 1) =

Mathematics

2 answers:

White raven [17]3 years ago

4 0

3(n+1)=
3(4+1=
15

Hope this helps!

Luda [366]3 years ago

4 0

Answer: Your answer is 15!

Step-by-step explanation: First, you have to do 4 + 1, equaling 5. Next, 5 x 3, which is 15.

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Continue the pattern for 3 more numbers, 1, 3, 7, 13, 21,__, __, __.<br><br> What is the pattern??

statuscvo [17]

If you are doing a simple version, it goes up by 2, then 4, then 6, then 8. So the next ones are 31, 43, 57

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3 years ago

Lucy separates a number of paintbrushes, x, into 2 equally sized groups. She claims that this is the same as making a group 1/2

irakobra [83]

Answer:

x/2 = 1/2x

Step-by-step explanation:

# of paintbrushes/ 2 piles = 1/2 # of paintbrushes

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3 years ago

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The CPI value in January 2009 was 211.143; the CPI value in January 2014 was 233.916. What was the 5-year inflation rate between

inna [77]

Answer:

211.143 233.916 Inflation rate: fo. Application of the formula to the inputs: IR = ?? -$0.10. Sample problem: The CPI value in January 2009 was 211.143

Step-by-step explanation:

8 0

3 years ago

Mislabeled seafood In 2013 the environmental group Oceana (usa.oceana.org) analyzed 1215 samples of seafood purchased across the

Step2247 [10]

Using the z-distribution, it is found that the 95% confidence interval for the proportion of all seafood sold in the United States that is mislabeled or misidentified is (0.3036, 0.3564).

<h3>z-distribution interval:</h3>

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which z is the z-score that has a p-value of $\frac{1+\alpha}{2}$ .

For this problem:

1215 samples, hence $n = 1215$ .

33% was mislabeled or misidentified, hence $p = 0.33$ .

95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so $z = 1.96$ .

<h3>The lower limit of this interval is:</h3>

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.33 - 1.96\sqrt{\frac{0.33(0.67)}{1215}} = 0.3036$

<h3>The upper limit of this interval is:</h3>

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.33 + 1.96\sqrt{\frac{0.33(0.67)}{1215}} = 0.3564$

The 95% confidence interval for the proportion of all seafood sold in the United States that is mislabeled or misidentified is (0.3036, 0.3564).

You can learn more about the use of the z-distribution to build a confidence interval at brainly.com/question/25730047

4 0

3 years ago

Please help i beg plsssssssssz

Minchanka [31]

Answer:

5/2=20/8=35/14=125/50

4 0

3 years ago

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