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1 year ago

Is (0, 2) a solution to the equation y = 7x?

Mathematics

2 answers:

o-na [289]1 year ago

6 0

Answer:

Step-by-step explanation:

when x is 0, y is also 0. the ordered pair for this is (0,0). y cannot be 0 and 2 at the same time, so (0,2) is not a solution,

jeka941 year ago

4 0

Answer:

Step-by-step explanation:

x = 0

y = 2

y = 7x

2 = 7(0)

2 ≠ 0

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Find the area of the shaded portion in the circle.

Anastasy [175]

See the attached picture to better understand the problem

we know that
the area of the shaded portion in the circle=area 1+area 2

step 1
find area 1
area 1=area of semi circle
area 1=pi*r²/2
diameter=12 units
radius r=12/2----> 6 units

r=6 units
area 1=pi*6²/2----> 56.52 units²

step 2
find the area 2

area 2=area sector CADC-area triangle ACD

in the right triangle ABC
BC=3
AC=r-----> AC=6
AB=?
applying Pythagoras Theorem
AC²=AB²+BC²------> AB²=6²-3²-----> AB²=27----> AB=3√3 units
area triangle ACD=b*h/2
b=2*3√3---> 6√3 units
h=3 units
area triangle ACD=6√3*3/2----> 9√3 units²-----> 15.59 units²

find the angle ACB
tan ∠ACB=AB/BC-----> 3√3/3---> √3
∠ACB=arc tan (√3)-----> 60°

the central angle ACD=60*2-----> 120°

area of sector CADC=(120/360)*pi*r²----> (120/360)*pi*6²---> 37.68 units²
area 2=area sector CADC-area triangle ACD
area 2=37.68-15.59----> 22.09 units²

step 3
the area of the shaded portion in the circle=area 1+area 2
the area of the shaded portion in the circle= 56.52 +22.09----> 78.61 units²

the answer is
the area of the shaded portion is 78.61 units²

6 0

2 years ago

Read 2 more answers

80 POINTS !!

givi [52]

Answer:

-4x^3+x+7

Step-by-step explanation:

6 0

3 years ago

(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

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

Let's approximate the distribution with a normal distribution with mean 170 and standard deviation 5.05

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

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

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