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BabaBlast [244]
2 years ago
14

What is the base 5 representation of the number 4658

Mathematics
1 answer:
Liula [17]2 years ago
6 0

Answer: There are five "digits": 0,1, 2, 3, 4

Step-by-step explanation: The base is five. There are five "digits": 0,1, 2, 3, 4 . Positions correspond to integer powers of five, starting with power 0 at the rightmost digit, and increasing right to left. The digit placed at a position shows how many times that power of five is included in the number.

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Two companies rent boats by the hour. The total cost, in dollars, c, depends on the number of hours, h.
Maurinko [17]

Answer:

A(35)

B(35_20=15

Step-by-step explanation:

same as all above

3 0
2 years ago
Allen�s hummingbird (Selasphorus sasin ) has been studied by zoologist Bill Alther A small group of 15 Allen� s hummingbirds has
Bogdan [553]

Answer:

1) 80% CI: [3.04; 3.26]gr

d= 0.11

2) n= 28 hummingbirds

Step-by-step explanation:

Hello!

The study variable of this experiment is:

X: the weight of a hummingbird. (gr)

And it has a normal distribution, symbolically: X~N(μ;σ²)

And (I hope I got it correctly) its population standard deviation is σ= 0.33

There was a sample of n= 15 hummingbirds taken, its sample mean X[bar]= 3.15 gr

1) You need to construct an 80% Confidence Interval for the population mean of the hummingbird's weight.

Since the study variable has a normal distribution, you can use either the standard normal distribution or the Student's t distribution. Both are useful to estimate the population mean. Since the population standard variance is known, the best choice is the Standard normal.

Z= <u> X[bar] - μ </u>~ N(0;1)

       σ/√n

The formula for the interval is:

X[bar] ± Z_{1- \alpha /2} * (σ/√n)

Z_{1- \alpha /2}= Z_{0.90} = 1.28

3.15 ± 1.28 * (0.33/√15)

[3.04; 3.26]gr

The margin of error (d) of a confidence interval is hal its amplitude (a)

a= Upper bond - Lower bond

d= (Upper bond - Lower bond)/2

d= \frac{(3.26-3.04)}{2} = 0.11

2) You need to calculate a sample size for a 80% Confidence interval for the average weight of the hummingbirds with a margin of error of d= 0.08

As I said before, the margin of error is half the amplitude of the interval, the formula you use to estimate the population mean has the following structure:

"point estamator" ± "margin of error"

Then the margin of error is:

d= Z_{1- \alpha /2} * (σ/√n)

Now what you have to do is rewrite the formula based on the sample size

d= Z_{1- \alpha /2} * (σ/√n)

\frac{d}{Z_{1- \alpha /2}}= σ/√n

√n * \frac{d}{Z_{1- \alpha /2}}= σ

√n = σ * \frac{Z_1- \alpha /2}{d}

n = (σ * \frac{Z_1- \alpha /2}{d})²

n=  (0.33 * \frac{1.28}{0.08})²

n= 27.8784 ≅ 28 hummingbirds.

I hope it helps!

4 0
3 years ago
Change 95.61 to degrees, minutes, and seconds.
12345 [234]
The answers are 95 36 36
6 0
3 years ago
Read 2 more answers
A recent survey by the cancer society has shown that the probability that someone is a smoker is P(S)=0.29. They have also deter
GuDViN [60]

<u>Answer:</u>

The correct answer option is P (S∩LC) = 0.16.

<u>Step-by-step explanation:</u>

It is known that the probability if someone is a smoker is P(S)=0.29 and the probability that someone has lung cancer, given that they are also smoker is P(LC|S)=0.552.

So using the above information, we are to find the probability hat a random person is a smoker and has lung cancer P(S∩LC).

P (LC|S) = P (S∩LC) / P (S)

Substituting the given values to get:

0.552 = P(S∩LC) / 0.29

P (S∩LC) = 0.552 × 0.29 = 0.16

5 0
3 years ago
HELP PLEASE <br> I DONT UNDERSTAND THIS AT ALL <br> slope =
Elanso [62]

Answer:

look at the picture i have sent

5 0
3 years ago
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