Transform the equation x-3y=-15 to express it in slope-intercept form

1 answer:

3 0

X -3y = -15

Subtract 'x' from each side:

-3y = -x - 15

Divide each side by -3 :

<em>y = 1/3 x + 5</em>

This is 'slope-intercept' form.

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A kitten’s mass at birth was 0.09 kilogram. The kitten gained approximately 0.084 kilogram each week. After how many weeks is th

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A kitten’s mass at birth was 0.09 kilogram. The kitten gained approximately 0.084 kilogram each week. After how many weeks is the kitten’s mass 1.098 kilograms?

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Given z1 = 2 +StartRoot 3 EndRoot i and z2 = 1 – StartRoot 3 EndRoot i, what is the sum of z1 and z2?

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Answer:

z1 + z2 = 3

Step-by-step explanation:

Since we are given z1 = 2 + √(3)i and z2 = 1 – √(3)i. The sum of z1 + z2 would be:

(2 + √(3)i) + (1 – √(3)i) = 2 + √(3)i + 1 – √(3)i = 2 + 1 + √(3)i – √(3)i = 3

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Polly's Polls asked 1850 second-year college students if they still had their original major. According to the colleges, 65% of

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Answer:

The probability that Polly's Sample will give a result within 1% of the value 65% is 0.6424

Step-by-step explanation:

The variable that assigns the value 1 if a person had its original major and 0 otherwise is a Bernoulli variable with paramenter 0.65. Since she asked the question to 1850 people, then the number of students that will have their original major is a Binomial random variable with parameters n = 1850, p = 0.65.

Since the sample is large enough, we can use the Central Limit Theorem to approximate that random variable to a Normal random variable, which we will denote X.

The parameters of X are determined with the mean and standard deviation of the Binomal that we are approximating. The mean is np = 1850*0.65 = 1202.5, and the standard deviation is √np(1-p) = √(1202.5*0.35) = 20.5152.

We want to know the probability that X is between 0.64*1850 = 1184 and 0.66*1850 = 1221 (that is, the percentage is between 64 and 66). In order to calculate this, we standarize X so that we can work with a standard normal random variable W ≈ N(0,1). The standarization is obtained by substracting the mean from X and dividing the result by the standard deviation, in other words

$W = \frac{X-\lambda}{\sigma} = \frac{X-1202.5}{20.5152}$

The values of the cummulative function of the standard normal variable W, which we will denote $\phi$ are tabulated and they can be found in the attached file.

Now, we are ready to compute the probability that X is between 1184 and 1221. Remember that, since the standard random variable is symmetric through 0, then \phi(-z) = 1-\phi(z) for each positive value z.

$P(1184 < X < 1221) = P(\frac{1184-1202.5}{20.5152} < \frac{X-1202.5}{20.5152} < \frac{1221-1202.5}{20.5152})\\ = P(-0.9018 < W < 0.9018) = \phi(0.9018) - \phi(-0.9018) = \phi(0.9018)-(1-\phi(0.9018))\\ = 2\phi(0.9018)-1 = 2*0.8212-1 = 0.6424$