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3 years ago

Help please WORTH 15 POINTS!

Mathematics

2 answers:

tino4ka555 [31]3 years ago

7 0

The correct answer is 22

lord [1]3 years ago

6 0

Answer:

Step-by-step explanation:

I think it is 22

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I need help, WILL GIVE BRAINLIEST IF RIGHT. i need it in percents

atroni [7]

Answer:

2.) 33.33%

3.) 50%

Step-by-step explanation:

There is only one 5 and only one 2. There are 6 sides. The fraction is 2/6, and as a percent it's 33.33%.

There are 3 even numbers and three odd numbers. The chance of getting an even then an odd would be 3/6, or 50%.

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3 years ago

Solve the equation: 12+3/4x=14

Alinara [238K]

Answer:

x=2 and 2/3

Step-by-step explanation:

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3 years ago

Read 2 more answers

(43 points) In the US, 85% of the population has Rh positive blood. Suppose we take a random sample of 6 persons and let Y denot

VladimirAG [237]

Answer:

a) Binomial distribution with parameters p=0.85 q=0.15 n=6

b) 62.29%

c) 2.38%

d) See explanation below

Step-by-step explanation:

We could model this situation with a binomial distribution

$P(6;k)=\binom{6}{k}p^kq^{6-k}$

where P(6;k) is the probability of finding exactly k people out of 6 with Rh positive, p is the probability of finding one person with Rh positive and q=(1-p) the probability of finding a person with no Rh.

$\bf P(Y=k)=\binom{6}{k}(0.85)^k(0.15)^{6-k}$

The probability that Y is less than 6 is

P(Y=0)+P(Y=1)+...+P(Y=5)

Let's compute each of these terms

$P(Y=0)=P(6;0)=\binom{6}{0}(0.85)^0(0.15)^{6}=1.139*10^{-5}$

$P(Y=1)=P(6;1)=\binom{6}{1}(0.85)^1(0.15)^{5}=0.0000387281$

$P(Y=2)=P(6;2)=\binom{6}{2}(0.85)^2(0.15)^{4}=0.005486484$

$P(Y=3)=P(6;3)=\binom{6}{3}(0.85)^3(0.15)^{3}=0.041453438$

$P(Y=4)=P(6;4)=\binom{6}{4}(0.85)^4(0.15)^{2}=0.176177109$

$P(Y=5)=P(6;5)=\binom{6}{5}(0.85)^5(0.15)^{1}=0.399334781$

and adding up these values we have that the probability that Y is less than 6 is

$\sum_{i=1}^{5}P(Y=i)=0.622850484\approx 0.6229=62.29\%$

In this case is a binomial distribution with n=200 instead of 6.

p and q remain the same.

The mean of this sample would be 85% of 200 = 170.

In a binomial distribution, the standard deviation is

$s = \sqrt{npq}$

In this case

$\sqrt{200(0.85)(0.15)}=5.05$

<em>Let's approximate the distribution with a normal distribution with mean 170 and standard deviation 5.05</em>

So, the approximate probability that there are fewer than 160 persons with Rh positive blood in a sample of 200 would be the area under the normal curve to the left of 160

(see picture attached)

We can compute that area with a computer and find it is

0.0238 or 2.38%

d)<em> In order to approximate a binomial distribution with a normal distribution we need a large sample like the one taken in c).</em>

In general, we can do this if the sample of size n the following inequalities hold:

$np\geq 5 \;and\;nq \geq 5$

in our case np = 200*0.85 = 170 and nq = 200*0.15 = 30

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3 years ago

When Isabel began her book-selling business, she stored her inventory in her garage. Now that her business has grown, she wants

nekit [7.7K]

Answer:

The answer is 17.

Step-by-step explanation:

3 0

2 years ago

Find the volume of the wedge-shaped region contained in the cylinder x2 + y2 = 49, bounded above by the plane z = x and below by

fiasKO [112]

$\displaystyle\iiint_R\mathrm dV=\int_{y=-7}^{y=7}\int_{x=-\sqrt{49-y^2}}^{x=0}\int_{z=x}^{z=0}\mathrm dz\,\mathrm dx\,\mathrm dy$

Converting to cylindrical coordinates, the integral is equivalent to

$\displaystyle\iiint_R\mathrm dV=\int_{\theta=\pi/2}^{\theta=3\pi/2}\int_{r=0}^{r=7}\int_{z=r\cos\theta}^{z=0}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle\int_{\theta=\pi/2}^{\theta=3\pi/2}\int_{r=0}^{r=7}-r^2\cos\theta\,\mathrm dr\,\mathrm d\theta$
$=-\displaystyle\left(\int_{\theta=\pi/2}^{3\pi/2}\cos\theta\,\mathrm d\theta\right)\left(\int_{r=0}^{r=7}r^2\,\mathrm dr\right)$
$=\dfrac{2\times7^3}3=\dfrac{686}3$

4 0

3 years ago