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Sunny_sXe [5.5K]
2 years ago
14

Please help me and please answer correctly. last try, please

Mathematics
2 answers:
Digiron [165]2 years ago
7 0

Answer:

B

Step-by-step explanation:

The y-intercept is 0,2 so that rules out C and D.

We then see that A cannot be true so it must be B.

P.S. I'd love to have brainiest plz!

Gnom [1K]2 years ago
7 0

Answer:

Step-by-step explanation:

It’s A

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The answer is 4x^3 + 4x + C

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If the standard deviation of an exam is 5, the z-score us 1.95 and the mean is 80; what is the actual test score? (Round the ans
skad [1K]

If the standard deviation of an exam is 5, the z-score us 1.95 and the mean is 80, the actual test score is; 89.75

<h3>How to solve z-score problems?</h3>

We are given;

Standard deviation; s = 5

z-score = 1.95

Mean = 80

Formula for z-score is;

z = (x' - μ)/σ

Thus;

1.95 = (x' - 80)/5

1.95 * 5 = (x' - 80)

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2 years ago
Let the equation of a line be described by Equation A:
umka2103 [35]
10y - 5x = 40   Add 5x to both sides
       10y = 5x + 40   Divide both sides by 10
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The y-intercept is 4 and the slope of the line is \frac{1}{2}.

You can find these by comparing your equation to the equation y = mx + b, where m is the slope of the line and b is the y-intercept.
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4 years ago
A certain test preparation course is designed to help students improve their scores on the LSAT exam. A mock exam is given at th
OlgaM077 [116]

Answer:

Interval [16.34 , 21.43]

Step-by-step explanation:

First step. <u>Calculate the mean</u>

\bar X=\frac{(23+18+23+12+13+23)}{6}=18.666

Second step. <u>Calculate the standard deviation</u>

\sigma =\sqrt{\frac{(23-18.666)^2+(18-18.666)^2+(23-18.666)^2+(12-18.666)^2+(13-18.666)^2+(23-18.666)^2}{6}}

\sigma=\sqrt\frac{18.783+0.443+18.783+44.435+5.666+18.783}{6}

\sigma=\sqrt{17.815}=4.22

As the number of data is less than 30, we must use the t-table to find the interval of confidence.

We have 6 observations, our level of confidence DF is then 6-1=5 and we want our area A to be 80% (0.08).  

We must then choose t = 1.476 (see attachment)

Now, we use the formula that gives us the end points of the required interval

\bar X \pm t\frac{\sigma}{\sqrt n}

where n is the number of observations.

The extremes of the interval are then, rounded to the nearest hundreth, 16.34 and 21.43

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3 years ago
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