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djverab [1.8K]
4 years ago
8

Domestic bees make their honeycomb by starting with a single hexagonal cell, then forming ring after ring of hexagonal cells aro

und the initial cell, as shown. The number of cells in successive rings forms an arithmetic sequence..
A-Write a rule for the number of cells in the
ring.


b. How many cells are in the honeycomb after the ninth ring is formed?

Mathematics
1 answer:
maw [93]4 years ago
7 0

Answer:

aₙ= 6 n

Step-by-step explanation:

A honey comb cell contain 6 cells

a₁ = 6

d=6

aₙ = 6 + (n-1)6

   = 6 n

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Gillian and her friends are playing a card game using a standard deck, but without the aces. In the deck, there are 36 numbered
Nostrana [21]

Answer:

30

Step-by-step explanation:

To find the best prediction for the amount of numbered cards that will be drawn during the game, first find the probability of drawing a numbered card. There are 36 numbered cards and 48 total cards. So, the probability of drawing a numbered card is  or .

Now, multiply the probability by the number of times a card will be drawn from the deck. Since there are 10 rounds and 4 players, a card will be drawn from the deck 10 × 4, or 40, times. So, multiply  by 40.

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3 years ago
A random sample of 10 parking meters in a beach community showed the following incomes for a day. Assume the incomes are normall
Vlad1618 [11]

Answer: (2.54,6.86)

Step-by-step explanation:

Given : A random sample of 10 parking meters in a beach community showed the following incomes for a day.

We assume the incomes are normally distributed.

Mean income : \mu=\dfrac{\sum^{10}_{i=1}x_i}{n}=\dfrac{47}{10}=4.7

Standard deviation : \sigma=\sqrt{\dfrac{\sum^{10}_{i=1}{(x_i-\mu)^2}}{n}}

=\sqrt{\dfrac{(1.1)^2+(0.2)^2+(1.9)^2+(1.6)^2+(2.1)^2+(0.5)^2+(2.05)^2+(0.45)^2+(3.3)^2+(1.7)^2}{10}}

=\dfrac{30.265}{10}=3.0265

The confidence interval for the population mean (for sample size <30) is given by :-

\mu\ \pm t_{n-1, \alpha/2}\times\dfrac{\sigma}{\sqrt{n}}

Given significance level : \alpha=1-0.95=0.05

Critical value : t_{n-1,\alpha/2}=t_{9,0.025}=2.262

We assume that the population is normally distributed.

Now, the 95% confidence interval for the true mean will be :-

4.7\ \pm\ 2.262\times\dfrac{3.0265}{\sqrt{10}} \\\\\approx4.7\pm2.16=(4.7-2.16\ ,\ 4.7+2.16)=(2.54,\ 6.86)

Hence, 95% confidence interval for the true mean= (2.54,6.86)

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