You are a restaurant manager and are responsible for ordering ingredients for the restaurant at supplier number one you can purc
1 answer:
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Answer:
yard
Step-by-step explanation:
Length of string Mrs. Ruiz has = yard
Mrs. Ruiz needs to hang 3 banners using the string.
Having known that each banner uses same length of string to be hung we can divide total length of sting into 3 equal parts.
So, we can divide yard by 3 as
⇒
Changing the division sing to multiplication by taking reciprocal of divisor.
⇒
We can now multiply numerators and denominators.
⇒ yard
∴ Mrs. Ruiz must used yard of string for each banner.
Answer:
Step-by-step explanation:
c. never.
Answer:
a)
<em>The probability that a single score drawn at random will be greater than 110 </em>
<em>P( X > 110) = 0.1587</em>
<em>b) </em>
<em>The probability that a sample of 25 scores will have a mean greater than 105</em>
<em> P( x> 105) = 0.0062</em>
<em>c) </em>
<em>The probability that a sample of 64 scores will have a mean greater than 105</em>
<em> P( x⁻> 105) = 0.002</em>
<em></em>
<em>d) </em>
<em> The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105</em>
<em> P( 95 ≤ X≤ 105) = 0.9544</em>
<em></em>
Step-by-step explanation:
<u><em>a)</em></u>
Given mean of the Normal distribution 'μ' = 100
Given standard deviation of the Normal distribution 'σ' = 10
a)
Let 'X' be the random variable of the Normal distribution
let 'X' = 110
<em>The probability that a single score drawn at random will be greater than 110</em>
<em>P( X > 110) = P( Z >1)</em>
= 1 - P( Z < 1)
= 1 - ( 0.5 +A(1))
= 0.5 - A(1)
= 0.5 -0.3413
= 0.1587
b)
let 'X' = 105
<em>The probability that a single score drawn at random will be greater than 110</em>
<em> P( x> 105) = P( z > 2.5)</em>
<em> = 1 - P( Z< 2.5)</em>
<em> = 1 - ( 0.5 + A( 2.5))</em>
<em> = 0.5 - A ( 2.5)</em>
<em> = 0.5 - 0.4938</em>
<em> = 0.0062</em>
<em>The probability that a single score drawn at random will be greater than 105</em>
<em> P( x> 105) = 0.0062</em>
<em>c) </em>
let 'X' = 105
<em>The probability that a single score drawn at random will have a mean greater than 105</em>
<em> P( x> 105) = P( z > 4)</em>
<em> = 1 - P( Z< 4)</em>
<em> = 1 - ( 0.5 + A( 4))</em>
<em> = 0.5 - A ( 4)</em>
<em> = 0.5 - 0.498</em>
<em> = 0.002</em>
<em> The probability that a sample of 64 scores will have a mean greater than 105</em>
<em> P( x⁻> 105) = 0.002</em>
<em>d) </em>
<em>Let x₁ = 95</em>
<em>Let x₂ = 105</em>
The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105
P( 95 ≤ X≤ 105) = P( -2≤z≤2)
= P(z≤2) - P(z≤-2)
= 0.5 + A( 2) - ( 0.5 - A( -2))
= A( 2) + A(-2) (∵A(-2) =A(2)
= A( 2) + A(2)
= 2 × A(2)
= 2×0.4772
= 0.9544
<em> The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105</em>
<em> P( 95 ≤ X≤ 105) = 0.9544</em>
<em> </em>