Answer:
D. (1/4, -2)
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
<u>Algebra I</u>
- Solving systems of equations using substitution/elimination
Step-by-step explanation:
<u>Step 1: Define Systems</u>
y = -8x
4x - y = 3
<u>Step 2: Solve for </u><em><u>x</u></em>
<em>Substitution</em>
- Substitute in <em>y</em>: 4x - (-8x) = 3
- Simplify: 4x + 8x = 3
- Combine like terms: 12x = 3
- Isolate <em>x</em>: x = 3/12
- Simplify: x = 1/4
<u>Step 3: Solve for </u><em><u>y</u></em>
- Define equation: y = -8x
- Substitute in <em>x</em>: y = -8(1/4)
- Multiply: y = -2
Answer:
0.281 = 28.1% probability a given player averaged less than 190.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
A bowling leagues mean score is 197 with a standard deviation of 12.
This means that 
What is the probability a given player averaged less than 190?
This is the p-value of Z when X = 190.



has a p-value of 0.281.
0.281 = 28.1% probability a given player averaged less than 190.
Answer:
(12)3=36
Step-by-step explanation:
i don't know.. I think this is what the question is trying to say XD
Answer:
4.2
Step-by-step explanation:
simplify 12 and 20 to 6 and 10 then divide 6 by 10 and multiply by 7 to get the answer.
Answer:
96
Step-by-step explanation:
From the given information:
At 95% Confidence interval level,Level of significance
0.05, the value of Z from the standard normal tables = 1.96
Margin of Error = 0.10
Let assume that the estimated proportion = 0.5
therefore; the sample size n can be determined by using the formula: 


n = 96.04
n
96