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VladimirAG [237]
2 years ago
12

A card is drawn from a standard deck of 52 playing cards. Find the probability that the card is a four or a queen

Mathematics
1 answer:
laiz [17]2 years ago
3 0
8/52 have a great day
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The sum of three consecutive even integer is 42. What is the value of the second largest integer
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5 0
2 years ago
Sheila has x 20 dollar bills, and Tamsin has y 10 dollar bills. Sheila has a total of $50 more than Tamsin, but Tamsin has 1 mor
natka813 [3]

Answer:

Sheila have 6 bills

Step-by-step explanation:

Suppose Sheila has x '20' dollar bills, so the total cost of her bill = $20x

Suppose Tamsin has y '10' dollar bills, so the total cost of her bill = $10y

<h3>Equation 1</h3>

<em>Sheila has a total of $50 more than Tamsin</em>

<h2><em>10y + 50 = 20x</em></h2><h2 /><h3>Equation 2</h3>

<em>Tamsin has 1 more bill than Sheila.</em>

<h2>y = x + 1 </h2><h2 />

<em>Now solve them by putting value of x in equation </em>

10( x + 1 ) + 50 = 20x

10x + 10 + 50 = 20x

60 = 20x - 10x

10x = 60

x = 60/10

x = 6

and

y = 7

<em> </em>

6 0
3 years ago
An insurance company selected samples of clients under 18 years of age and over 18 and recorded the number of accidents they had
Stolb23 [73]

Answer:

Q1 z(s) is in the rejection region for H₀ ; we reject H₀. We can´t support the that means have no difference

Q2  CI 95 %  =  (  0,056 ;  0,164 )

Step-by-step explanation:

Sample information for people under 18

n₁  =  500

x₁ =  180

p₁  =  180/ 500    p₁  =  0,36    then  q₁  =  1 -  p₁     q₁ =  0,64

Sample information for people over 18

n₂  =  600

x₂  =  150

p₂  =  150 / 600   p₂ =  0,25   then   q₂  =  1 - p₂   q₂ =  1 - 0,25   q₂ = 0,75

Hypothesis Test

Null hypothesis                        H₀              p₁  =  p₂

Alternative Hypothesis           Hₐ              p₁  ≠  p₂

The alternative hypothesis indicates that the test is a two-tail test.

We will use the approximation to normal distribution of the binomial distribution according to the sizes of both samples.

Testin at CI =  95 %    significance level is  α = 5 %   α  =  0,05  and

α/ 2  =  0,025   z (c) for that α  is from z-table:

z(c) = 1,96

To calculate   z(s)

z(s)  =  ( p₁  -   p₂ ) / EED

EED = √(p₁*q₁)n₁  +  (p₂*q₂)/n₂

EED = √( 0,36*0.64)/500  +  (0,25*0,75)/600

EED = √0,00046  +  0,0003125

EED = 0,028

( p₁  -  p₂  )  =  0,36  -  0,25  = 0,11

Then

z(s)  =  0,11 / 0,028

z(s) = 3,93

Comparing  z(s) and  z (c)    z(s) > z(c)

z(s) is in the rejection region for H₀ ; we reject H₀. We can´t support the idea of equals means

Q2  CI  95 %   =  (  p₁  -  p₂  ) ±  z(c) * EED

CI 95%  =  ( 0,11   ±  1,96 * 0,028 )

CI 95%  = (  0,11  ±  0,054 )

CI 95 %  =  (  0,056 ;  0,164 )

8 0
2 years ago
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