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3 years ago

2 concentric circles have radii 2cm and 3cm respectively, calculate the ratio of their areas

1 answer:

6 0

Answer:For the first circle the radius is 2cm and for the second circle the radius is 3cm,by taking the ratio of their areas:

πr^2:πR^2

π cancels out and we are left with only r^2:R^2

r=2cm and R=3cm

Therefore r^2:R^2=2^2:3^2

=4:9

The answer is 4:9

Step-by-step explanation:

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Simplify 3x^2x 4x^5 please help

Firdavs [7]

Answer:

$12x^7$

Step-by-step explanation:

$3x^2 \times 4x^5$

$3 \times x^2 \times 4 \times x^5$

$3 \times 4 \times x^{2+5}$

$12 \times x^7$

8 0

3 years ago

A random variable is not normally distributed, but it is mound shaped. It has a mean of 14 and a standard deviation of 3. If you

Over [174]

Answer:

Step-by-step explanation:

from the question,

the mean 14

the standard deviation is 3

and sample size is 10.

since the n which is the sample size is 10, then the distribution is mound shaped.

why?

this is due to the fact that the random variable from which we took the sample is mound shaped.

The sampling distribution of the mean is normally distributed although the question says the random variable is not normally distributed. so the shape is bell shaped and normally distributed.

the standard deviation of the mean is

3/√10

= 0.948

5 0

4 years ago

A bag contains red marbles, white marbles, and blue marbles. Randomly choose two marbles, one at a time, and without replacement

dsp73

Answer:

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

$P(Same) = \frac{67}{210}$

Step-by-step explanation:

Given (Omitted from the question)

$Red = 7$

$White = 9$

$Blue = 5$

Solving (a): $P(First\ White\ and\ Second\ Blue)$

This is calculated using:

$P(First\ White\ and\ Second\ Blue) = P(White) * P(Blue)$

$P(First\ White\ and\ Second\ Blue) = \frac{n(White)}{Total} * \frac{n(Blue)}{Total - 1}$

<em>We used Total - 1 because it is a probability without replacement</em>

So, we have:

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{21 - 1}$

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{20}$

$P(First\ White\ and\ Second\ Blue) = \frac{9*5}{21*20}$

$P(First\ White\ and\ Second\ Blue) = \frac{45}{420}$

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

Solving (b) $P(Same)$

This is calculated as:

$P(Same) = P(First\ Blue\ and Second\ Blue)\$ $or\ P(First\ Red\ and Second\ Red)\ or\ P(First\ White\ and Second\ White)$

$P(Same) = (\frac{n(Blue)}{Total} * \frac{n(Blue)-1}{Total-1})+(\frac{n(Red)}{Total} * \frac{n(Red)-1}{Total-1})+(\frac{n(White)}{Total} * \frac{n(White)-1}{Total-1})$

$P(Same) = (\frac{5}{21} * \frac{4}{20})+(\frac{7}{21} * \frac{6}{20})+(\frac{9}{21} * \frac{8}{20})$

$P(Same) = \frac{20}{420}+\frac{42}{420} +\frac{72}{420}$

$P(Same) = \frac{20+42+72}{420}$

$P(Same) = \frac{134}{420}$

$P(Same) = \frac{67}{210}$

6 0

3 years ago

Find the missing factor A that makes the equality true.

olganol [36]

I’m kinda confused but I’d go with:

Expand the brackets

-2xA = 82

/-2 on both sides

xA = -41

/x on both sides

A = -41/x

4 0

4 years ago

Read 2 more answers

Aly is looking for a new car. She has test-driven two cars but can only purchase one. The probability that she will purchase car

bixtya [17]

The probability of buying A or B is P(A∪B) = 0.47 + 0.35

P(A∪B) = 0.82
But the probability of NOT buying either is :
P(A'∪B') = 1 - 0.82 = 0.18

5 0

3 years ago

Read 2 more answers

2 concentric circles have radii 2cm and 3cm respectively, calculate the ratio of their areas​

2 concentric circles have radii 2cm and 3cm respectively, calculate the ratio of their areas