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gtnhenbr [62]
2 years ago
13

Please answer all parts as I know the answers but need the work to go with them. Thus, I believe the above answers are correct.

Thank you!
Mathematics
1 answer:
vampirchik [111]2 years ago
8 0

As the sample size n is less than 30 normal distribution is used.

The values of x are converted into z by using the formula z= x-u/ s and then the z values are found out from the table.

The limits are found by using the formula x±σz or x±sz where s= σ

As the sample size is 10 which is less than 30 the normal distribution is used.

The probability of x< 2.59 is 0.3446

The probability 2.60<X <2.63 is 0.9484

So lower and upper limits are 2.607 and 2.612

Part A

As the sample size is 10 which is less than 30 the normal distribution is used.

Part B

For given value of x= 2.59 z is obtained =0.4

x= 2.59

z= x-u/ s

z= 2.59-2.61/0.05

z= -0.02/0.05

z=- 0.4

P (X<2.59) = P(-0.4 <Z<0) = 0.5 -0.1554= 0.3446

The probability of x< 2.59 is 0.3446

Part C

For two given values of x= 2.60 and 2.63 z is obtained as =0.2 and 0.4

x1= 2.60

z1= x-u/ s

z= 2.60-2.61/0.05

z= -0.01/0.05

z=- 0.2

x2= 2.63

z2= x-u/ s

z= 2.63-2.61/0.05

z= 0.02/0.05

z= 0.4

P (2.60<X<2.63) = P(-0.2 <Z<0.4)

= P(-0.2 <Z<0)+ P(0 <Z<0.4)

=0.793 + 0.1554= 0.9484

The probability 2.60<X <2.63 is 0.9484

Part D"

p= 0.57

From the table z= 0.045

z= x-u/ s

zs= x-u

zs+u = x

x1= 0.045*0.05 +2.61= 2.61225

x2= 2.61- 0.00225= 2.60775

So lower and upper limits are 2.607 and 2.612

For further understanding of probability calculation click

brainly.com/question/25638875

#SPJ1

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. For each of these intervals, list all its elements or explain why it is empty. a) [a, a] b) [a, a) c) (a, a] d) (a, a) e) (a,
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Answer:

Elements are of the form

 (i) [a,a]=\{[x,y] : a\leq x\leq a, a\leq y\leq a; a\in \mathbb R\}

(ii) [a,b)=\{[x,y) :a\leq x

(iii)(a,a]=\{(x,y] :a

(iv)(a,a)=\{(x,y): a

(v) (a,b) where a>b=\{(x,y) : a>x>b,a>y>b;a>b,a,b \in \mathbb R\}

(vi) [a,b] where a>b=\{[x,y] : a\geq x\geq b,a\geq y\geq b;a>b,a,b \in \mathbb R\}

Step-by-step explanation:

Given intervals are,

(i) [a,a] (ii) [a,a) (iii) (a,a] (iv) (a,a) (v) (a,b) where a>b (vi)  [a,b] where a>b.

To show all its elements,

(i) [a,a]

Imply the set including aa from left as well as right side.

Its elements are of the form.

\{[a,a] : a\in \mathbb R\}=\{[0,0],[1, 1],[-1,-1],[2,2],[-2,-2],[3,3],[-3,-3],........\}

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if a\in \mathbb R then the set  [a,a] represents singleton sets, and singleton sets are empty so is [a,a].

(ii) [a,a)

This means given interval containing a by left and exclude a by right.

Its elements are of the form.

[ 1, 1),[-1,-1),[2,2),[-2,-2),[3,3),[-3,-3),........

Since there is a singleton element a of real numbers withis the set, this set is empty.

Because there is no increment so if a\in \mathbb R then the set  [a,a) represents singleton sets, and singleton sets are empty so is [a.a).

(iii) (a,a]

It means the interval not taking a by left and include a by right.

Its elements are of the form.

( 1, 1],(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if a\in \mathbb R then the set  (a,a] represents singleton sets, and singleton sets are empty so is (a,a].

(iv) (a,a)

Means given set excluding a by left as well as right.

Since there is a singleton element a of real numbers, this set is empty.

Its elements are of the form.

( 1, 1),(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........

Because there is no increment so if a\in \mathbb R then the set  (a,a) represents singleton sets, and singleton sets are empty, so is (a,a).

(v) (a,b) where a>b.

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which not take x and y by any side of the interval.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

(a,b)=\{(5,0),(5,1),(5,2).....\} e.t.c

So this set is connected and we know singletons are connected in \mathbb R. Hence given set is empty.

(vi) [a,b] where a\leq b.

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which include both x and y.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

[a,b]=\{[5,0],[5,1],[5,2].....\} e.t.c

So this set is connected and we know singletons are connected in \mathbb R. Hence given set is empty.

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3 years ago
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Answer:

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Step-by-step explanation:

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0.85*\$450 = \$382.5

Thus, during the 15% off sale, the dishwasher will cost $382.5.

Now, the employees of the discount appliance store receive an additional 20% off; therefore, the price of the dishwasher will be 100% - 20% = 80% of $382.5:

\dfrac{80\%}{100\%} *\$382.5

0.8*\$382.5=\$306

Therefore, the employee will purchase the dishwasher at a price of $306, which is choice D.

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4 years ago
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You must check out the other 3 data sets in the same manner.  Pick the set that is most closely illustrated by the diagram.

5 0
4 years ago
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