3. Solve: -2 + (2)- 3 - (-1)<br> A. 1 1<br> ola -

How do you use equivalent fractions to compare the fractions in each pair

Alenkasestr [34]

Firstly you need to to reduce to a common denominator by multiplYing the whole fraction. This way you have to multiply 4/5 by 2 (both numerator and denominator). so 4/5 = 8/10. And now you have 8/10 and 9/10. You compare only numerators. This way you have 8/10 <9/10. Same goes for 2/3 and 5/8. Only here you need to multiply both fractions. The common denominator here is 24. So you have to multiply 2/3 by 8 and 5/8 by 3. You now have to compare 16/24 and 15/24. 16/24>15/24.

8 0

3 years ago

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.

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