Answer:
2.28% probability that a person selected at random will have an IQ of 110 or greater
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean 
and standard deviation 
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
What is the probability that a person selected at random will have an IQ of 110 or greater?
This is 1 subtracted by the pvalue of Z when X = 110. So
has a pvalue of 0.9772
1 - 0.9772 = 0.0228
2.28% probability that a person selected at random will have an IQ of 110 or greater
Answer:
The answer is D 6
Step-by-step explanation:
Multiply 3x2 and it equals 6
The correct answer is: [B]: " 25 a²⁵ b²⁵ " .
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<span>Explanation:
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Given the expression:
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</span>→ " (−5a⁵b⁵)² (a³b³)⁵ " ; Simplify.
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Let us being by examining:
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→ "(−5a⁵b⁵)² " .
→ "(−5a⁵b⁵)² = (-5)² * (a⁵)² * (b⁵)² = (-5)(-5) * a⁽⁵ˣ²⁾ * b⁽⁵ˣ²⁾ = 25a⁽¹⁰⁾b⁽¹⁰⁾ ;
{Note the following properties of exponents:
(xy)ⁿ = xⁿ * yⁿ ;
(xᵃ)ᵇ = x⁽ᵃ * ᵇ) ;
(xᵃ) * (xᵇ) = x⁽ᵃ ⁺ ᵇ⁾ .}.
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Then, we examine:
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→ "(a³b³)⁵ " .
→ "(a³b³)⁵ = a⁽³ˣ⁵⁾b⁽³ˣ⁵⁾ = a⁽¹⁵⁾b⁽¹⁵⁾ .
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So: " (−5a⁵b⁵)² (a³b³)⁵ = (-5)a⁽¹⁰⁾b⁽¹⁰⁾ * a⁽¹⁵⁾b⁽¹⁵⁾ " ;
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Now, we simplify:
→ " 25a⁽¹⁰⁾b⁽¹⁰⁾ * a⁽¹⁵⁾b⁽¹⁵⁾ " ;
→ " 25a⁽¹⁰⁾b⁽¹⁰⁾ * a⁽¹⁵⁾b⁽¹⁵⁾ ;
= 25a⁽¹⁰⁾ a⁽¹⁵⁾b⁽¹⁰⁾ b⁽¹⁵⁾ ;
= 25a⁽¹⁰ ⁺¹⁵⁾ b⁽¹⁰⁺¹⁵⁾ ;
= 25a⁽²⁵⁾ b⁽²⁵⁾ ;
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→ which is: Answer choice: [B]: " 25 a²⁵ b²⁵ " .
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14x+7=-91
-7=-91-7
14x=-98
*14 *14
x= -1372
Answer:
x = 89
Step-by-step explanation:
106 - 17 = 89
