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3 years ago

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approximately 0.02% of north atlantic lobsters are born bright blue in color. Out of 5,000 North Atlantic lobsters, how many wou

ld you expect to be blue in color

2 answers:

BigorU [14]3 years ago

6 0

Find 0.02% of all of them
percent means parts out of 100
0.02%=0.02/100=0.002/10=0.0002/1=0.0002

5000 times 0.0002=1

I would expect 1 to be blue

Julli [10]3 years ago

3 0

5000 divided by 0.02 is 100. Only about 100 of the lobsters will be born blue in color. Hope I helped!

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At The Last Book Store, 4 books cost $28. How many books could you buy for $49?

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You divide 28 by 4 and get 7. Then you divide 49 by whatever you get.

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A circle with area \blue{9\pi}9πstart color #6495ed, 9, pi, end color #6495ed has a sector with a central angle of \purple{60^\c

Liono4ka [1.6K]

Answer:

The area of the sector is 3pi/2

Step-by-step explanation:

A sector is that part of a circle bounded by 2 radii and an arc

The circle has an area of 9 pi with a central angle of 60 degrees.

Now the area of the sector is pretty much straight forward to calculate. Since it is a circle, the total angle we have is 360.

Now the sector subtends an angle of 60 degrees at the centre. What this means is that the sector is exactly 1/6 of the circle. meaning that cutting the circle into 6 slices will give the sector.

Thus, the area of the sector is one-sixth the area of the circle.

The area of the sector is thus 1/6 * 9pi = 9pi/6 = 3pi/2 or 3/2 pi

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Which table represents the graph of a logarithmic function in the form y=log3x when b>1?

alex41 [277]

Answer:

<u><em>The satisfied table of the given function</em></u> $y = log_{b} (x)$ <u><em></em></u>

<em>x 1/8 1/4 1/2 1 2</em>

<em>y -3 -2 -1 0 1</em>

<em></em>

Step-by-step explanation:

<u><em>Explanation</em></u> :-

Given logarithmic function $y = log_{b} (x)$ if b >1

Given first table

i)

put x = $\frac{1}{8}$ given b > 1 so we can choose b = 2

$y = log_{2} (\frac{1}{8} )$

$y = log_{2} (2^{-3} )$

we will apply logarithmic formula

log x ⁿ = n log (x)

$y = log_{2} (2^{-3} ) = -3 log_{2} (2) = -3 (1) = -3$

<em>y = -3</em>

<em>ii)</em>

<em>put x = </em> $\frac{1}{4}$ <em> given b > 1 so we can choose b = 2</em>

<em></em> $y = log_{2} (\frac{1}{4} )$ <em></em>

<em></em> $y = log_{2} (2^{-2} )$ <em></em>

we will apply logarithmic formula

log x ⁿ = n log (x)

$y = log_{2} (2^{-2} ) = -2 log_{2} (2) = -2 (1) = -2$

<em>y = -2</em>

<em>iii) </em>

<em>put x = </em> $\frac{1}{2}$ <em> given b > 1 so we can choose b = 2</em>

<em></em> $y = log_{2} (\frac{1}{2} )$ <em></em>

$y = log_{2} (2^{-1} )$

<em>we will apply logarithmic formula </em>

<em>log x ⁿ = n log (x)</em>

$y = log_{2} (2^{-1} ) = -1 log_{2} (2) = - (1) = -1$

<em>y = -1</em>

<em>iv) </em>

<em>put x = 1 given b > 1 so we can choose b = 2</em>

<em></em> $y = log_{2} (1 )$ <em> = 0</em>

<em>y = 0</em>

<em>v) </em>

<em>put x = </em> $2$ <em> given b > 1 so we can choose b = 2</em>

$y = log_{2} (2 )$

<em>y = 1</em>

<em></em>

<u><em>Final answer:-</em></u>

<u><em>The satisfied table of the given function</em></u>

<em>x 1/8 1/4 1/2 1 2</em>

<em>y -3 -2 -1 0 1</em>

<em></em>

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3 years ago

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Answer:

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3 years ago

The following results come from two independent random samples taken of two populations.

photoshop1234 [79]

Answer:

$(a)\ \bar x_1 - \bar x_2 = 2.0$

$(b)\ CI =(1.0542,2.9458)$

$(c)\ CI = (0.8730,2.1270)$

Step-by-step explanation:

Given

$n_1 = 60$ $n_2 = 35$

$\bar x_1 = 13.6$ $\bar x_2 = 11.6$

$\sigma_1 = 2.1$ $\sigma_2 = 3$

Solving (a): Point estimate of difference of mean

This is calculated as: $\bar x_1 - \bar x_2$

$\bar x_1 - \bar x_2 = 13.6 - 11.6$

$\bar x_1 - \bar x_2 = 2.0$

Solving (b): 90% confidence interval

We have:

$c = 90\%$

$c = 0.90$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.90$

$\alpha = 0.10$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.10/2}$

$z_{\alpha/2} = z_{0.05}$

The z score is:

$z_{\alpha/2} = z_{0.05} =1.645$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.645 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.645 * \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.645 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.645 * \sqrt{0.3306}$

$2.0 \± 0.9458$

Split

$(2.0 - 0.9458) \to (2.0 + 0.9458)$

$(1.0542) \to (2.9458)$

Hence, the 90% confidence interval is:

$CI =(1.0542,2.9458)$

Solving (c): 95% confidence interval

We have:

$c = 95\%$

$c = 0.95$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.95$

$\alpha = 0.05$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.05/2}$

$z_{\alpha/2} = z_{0.025}$

The z score is:

$z_{\alpha/2} = z_{0.025} =1.96$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.96 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.96* \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.96 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.96* \sqrt{0.3306}$

$2.0 \± 1.1270$

Split

$(2.0 - 1.1270) \to (2.0 + 1.1270)$

$(0.8730) \to (2.1270)$

Hence, the 95% confidence interval is:

$CI = (0.8730,2.1270)$

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