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-BARSIC- [3]
3 years ago
14

Number cube experiment number on the cube number of times appeared 1 6 2 4 3 0 4 9 5 3 6 2

Mathematics
1 answer:
photoshop1234 [79]3 years ago
5 0
Whts the question for this ?
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Which of the following options have the same value as 62%, percent of 45?
brilliants [131]

Answer:

A and C

Step-by-step explanation:

A is correct because converting the percentage to decimal form will give you the value of that percentage, and 62% will be the same as 0.62 in decimal form.

C is the same in a less simplified form because 62/100 = 0.62, so it is the same as A when you convert it to decimal form, 0.62 · 45.

8 0
3 years ago
A research study uses 800 men under the age of 55. Suppose that 30% carry a marker on the male chromosome that indicates an incr
ss7ja [257]

Answer:

a) There is a 12.11% probability that exactly 1 man has the marker.

b) There is a 85.07% probability that more than 1 has the marker.

Step-by-step explanation:

There are only two possible outcomes: Either the men has the chromosome, or he hasn't. So we use the binomial probability distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And \pi is the probability of X happening.

In this problem, we have that:

30% carry a marker on the male chromosome that indicates an increased risk for high blood pressure, so \pi = 0.30

(a) If 10 men are selected randomly and tested for the marker, what is the probability that exactly 1 man has the marker?

10 men, so n = 10

We want to find P(X = 1). So:

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 1) = C_{10,1}.(0.30)^{1}.(0.7)^{9} = 0.1211

There is a 12.11% probability that exactly 1 man has the marker.

(b) If 10 men are selected randomly and tested for the marker, what is the probability that more than 1 has the marker?

That is P(X > 1)

We have that:

P(X \leq 1) + P(X > 1) = 1

P(X > 1) = 1 - P(X \leq 1)

We also have that:

P(X \leq 1) = P(X = 0) + P(X = 1)

P(X = 0) = C_{10,0}.(0.30)^{0}.(0.7)^{10} = 0.0282

So

P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0282 + 0.1211 = 0.1493

Finally

P(X > 1) = 1 - P(X \leq 1) = 1 - 0.1493 = 0.8507

There is a 85.07% probability that more than 1 has the marker.

3 0
3 years ago
(x+5) (x-6)<br> Please give me the answer
o-na [289]

Answer:

x2−x−30 is the solution

5 0
3 years ago
Read 2 more answers
What is the value of this expression when a=20 5a
Darina [25.2K]

Answer:

100

Step-by-step explanation:


Given :     a=20


To Find :  the value of  expression 5a.

Solution :

We are given the value of 'a' i.e. a=20

Now to find the value of given expression  5a


⇒5a


put a = 20 in the equation


⇒5*20


⇒100


⇒5a = 100


Thus , The value of expression 5a when a = 20 is 100

3 0
4 years ago
Read 2 more answers
Math HW, simplify radical expressions, show all work.
Ivahew [28]

Answer:

12.  7xy ∛(xy^2)

13.  3x^2|y| sqrt(7x)  

14. 5  x^4   y ^4 y ∛(y^2)

Step-by-step explanation:

12.  (343 x^4y^5) ^ 1/3

separate  

343^1/3  x^4 ^ 1/3    y^5 ^ 1/3

using the power of power we can multiply the powers

343 ^ 1/3  x^ 4/3  y ^ 5/3

simplify

7   x^ 4/3  y ^ 5/3

if the power is greater than 1 we can split it

7 x x^ 1/3  y y ^ 2/3

7xy ( xy^2 ) ^ 1/3

7xy ∛(xy^2)


13.  (189 x^5y^6)^ 1/2/(3y^4)^ 1/2

combine

( (189 x^5y^6)/(3y^4))^ 1/2

(189/3  * x^5 * y^6/y^4) ^ 1/2

when we divide exponents with the same base, we subtract the exponent

(63 * x^5 * y^2) ^ 1/2

split

63 ^ 1/2   x^ 5 ^ 1/2  y ^ 2 ^ 1/2

sqrt (63) * sqrt * (x^5) * sqrt(y^2)

sqrt(9*7) sqrt( x^4 *x)  * sqrt(y^2)

sqrt(9) * sqrt(7)  sqrt(x^4) sqrt(x)  sqrt(y^2)

we need to make sure to take the positive value of sqrt(y^2)

3sqrt(7)  x^2  sqrt(x)  | y|  

3x^2|y| sqrt(7x)  


14.  (625 x^17 y^16) ^1/3  / (5 x^5 y^2) ^ 1/3

combine

(625 x^17 y^16  / 5 x^5 y^2) ^ 1/3

(625/5 * x^17/x^5   * y^16/y^2) ^ 1/3

when we divide exponents with the same base, we subtract the exponent

(125 x^(17-5)  y^(16-2)) ^ 1/3

(125 x^12 y ^ 14) ^ 1/3

split

125 ^ 1/3   x^ 12 ^ 1/3    y^ 14 ^ 1/3

power to the power ( multiply the power)

125 ^ 1/3 x^ 4 y ^ 14/3

5  x^4 y ^ 14/3

if the power is greater than 1 we can split it

5  x^4   y ^4 y ^2/3

5  x^4   y ^4 y ∛(y^2)

7 0
3 years ago
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