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Svetach [21]
2 years ago
10

PLEASE HELP What is the rule for the sequence with the first four terms below?

Mathematics
1 answer:
ivolga24 [154]2 years ago
3 0

Answer:

A) [First try is correct for all U1 to U4]

Step-by-step explanation:

1. Trying each option

(let x = 1,2,3,4) since it goes from U1 to U4 (Geometric sequence)

A.) 10(-4/5)^x

10(-4/5)^1 = -8

10(-4/5)^2 = 6.4

10(-4/5)^3 = -5.12

10(-4/5)^4 = 4.096

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2 years ago
According to the manufacturer of the candy Skittles, 20% of the candy produced are red. If we take a random sample of 100 bags o
igomit [66]

Answer:

Probability that the proportion in our sample of red candies will be less than 20% is 0.5 .

Step-by-step explanation:

We are given that 20% of the candy produced are red. A random sample of 100 bags of Skittles is taken.

The distribution we can use here is;

               \frac{\hat p - p}{\sqrt{\frac{\hat p (1- \hat p)}{n} } } ~ N(0,1)

where, p = 0.20 and n = 100

Let \hat p = proportion of red candies in our sample

So, P(\hat p < 0.20) = P(\frac{\hat p - p}{\sqrt{\frac{\hat p (1- \hat p)}{n} } } < \frac{0.20 - 0.20}{\sqrt{\frac{0.2 (1- 0.2)}{100} } } ) = P(Z < 0) = 0.5

Therefore, probability that the proportion in our sample of red candies will be less than 20% is 0.5 .

5 0
3 years ago
Find the variance of this probability distribution. Round to two decimal places.​
attashe74 [19]

Answer:

Variance = 4.68

Step-by-step explanation:

The formula for the variance is:

\sigma^{2} =\frac{\Sigma(X- \mu)^{2}}{N} \\or \\ \sigma^{2} =\frac{\Sigma(X)^{2}}{N} -\mu^{2} \\

Where:

X: Values \\\mu: Mean \\N: Number\ of\ values

The mean can be calculated as each value multiplied by its probability

\mu = 0*0.4 + 1*0.3 + 2*0.1+3*0.15+ 4*0.05=1.15

\frac{\Sigma (X)^{2}}{N} =\frac{(0^{2}+1^{2}+2^{2}+3^{2}+4^{2})}{5} =6

Replacing the mean and the summatory of X:

\sigma^{2} = \frac{\Sigma(X)^{2}}{N} -\mu^{2} \\= 6 - 1.15^{2}\\= 4.6775

4 0
3 years ago
Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $10,
Andreyy89

Answer:

1) the planning value for the population standard deviation is 10,000

2)

a) Margin of error E = 500, n = 1536.64 ≈ 1537

b) Margin of error E = 200, n = 9604

c) Margin of error E = 100, n = 38416

3)

As we can see, sample size corresponding to margin of error of $100 is too large and may not be feasible.

Hence, I will not recommend trying to obtain the $100 margin of error in the present case.

Step-by-step explanation:

Given the data in the question;

1) Planning Value for the population standard deviation will be;

⇒ ( 50,000 - 10,000 ) / 4

= 40,000 / 4

σ = 10,000

Hence, the planning value for the population standard deviation is 10,000

2) how large a sample should be taken if the desired margin of error is;

we know that, n = [ (z_{\alpha /2 × σ ) / E ]²

given that confidence level = 95%, so z_{\alpha /2  = 1.96

Now,

a) Margin of error E = 500

n = [ (z_{\alpha /2 × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 500 ]²

n = [ 19600 / 500 ]²

n = 1536.64 ≈ 1537

b) Margin of error E = 200

n = [ (z_{\alpha /2 × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 200 ]²

n = [ 19600 / 200 ]²

n = 9604

c)  Margin of error E = 100

n = [ (z_{\alpha /2 × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 100 ]²

n = [ 19600 / 100 ]²

n = 38416

3) Would you recommend trying to obtain the $100 margin of error?

As we can see, sample size corresponding to margin of error of $100 is too large and may not be feasible.

Hence, I will not recommend trying to obtain the $100 margin of error in the present case.

7 0
2 years ago
What do they mean by this!?!?!? I didn't learn this yet.... I think .......... look at the picture
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It means to break down the number 20
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