1) What is the perimeter of the triangle shown in the drawing?

1 answer:

8 0

I believe all you do is add the numbers with the same coefficients so 2x + 5x + x (in front of x is understood to be a 1) So you get 8x. And 8 + 1 is 9 so the perimeter is B. 8x + 9

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What is the area of a rectangle with a length of (x - 5) and width of (x + 8)?

dlinn [17]

Answer:

a = x² + 3x - 40

Step-by-step explanation:

a = l * w

a = (x - 5)(x + 8)

a = x(x + 8) - 5(x + 8)

a = (x² + 8x) + (- 5x - 40)

a = x² + 3x - 40

3 0

3 years ago

Read 2 more answers

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

suter [353]

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.

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