Write the first five terms of each sequence described and identify the sequence as

Mathematics

1 answer:

Vedmedyk [2.9K]3 years ago

8 0

Answer: i think its 4 im not sure

Step-by-step explanation:

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Joe completed 7 pages of homework over 3 hours. Nina completed 5 pages over two hours. Who was completing the work at a faster p

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Answer:

Joe

Step-by-step explanation:

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What's the answer for each to get 30 points!

MAVERICK [17]

Answer: "greater than" for each of the 4 dropdown menus

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Explanation:

Divide each value in the table by 40

You should get:

Horse = 15/40 = 0.375
Cow = 12/40 = 0.300
Sheep = 14/40 = 0.350
Pig = 19/40 = 0.475

Those decimal results are the experimental (ie empirical) probabilities. Theoretically, we should get 1/4 = 0.250 for each sticker type assuming each sticker is likely to be chosen. As you can see, each decimal value shown above is larger than the theoretical target of 0.250, so each answer is "greater than"

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Here's another way to see why this is:

If we had 40 stickers total, and each animal has the same number of stickers, then we should have 40/4 = 10 stickers per animal type. But the table shows each frequency is above 10. So that must directly mean the empirical probability of picking any animal is greater than the theoretical probability.

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3 years ago

PLEASE HELP WILL MARK BRAINLIEST find the value of x

algol [13]

I think,

the value of x is 90°

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11) A marketing research company is estimating the average total compensation of CEOs in the service industry. Data were randoml

Illusion [34]

Answer:

The confidence interval would be more narrow if the confidence level were changed to 90%

Step-by-step explanation:

For a small sample size of n = 18 a pivotal quantity that we can use to form a confidence interval for $\mu$ is given by $T=\frac{\bar{X}-\mu}{S/\sqrt{n}}$ that has a t distribution with (n-1) degrees of freedom. We find a $100(1-\alpha)$ % confidence inverval using $P(-t_{\alpha/2}\leq T \leq t_{\alpha/2}) = 1-\alpha$ , where $t_{\alpha/2}$ is the t-value such that there is an area equal to $\alpha/2$ above this t-value and below the curve of the density of the t distribution with n-1 df. We find a 95% confidence interval with $P(-t_{0.025}\leq T\leq t_{0.025}) = 0.95$ and we find a 90% confidence interval with $P(-t_{0.05}\leq T\leq t_{0.05}) = 0.90$ . Because of $-t_{0.025} < -t_{0.05}$ and $t_{0.05} < t_{0.025}$ , the confidence interval would be more narrow if the confidence level were changed to 90%.