Simplify the expression <br> -7(3e-2f+4)+6e-2

Brayden and Aubrey are both driving along the same highway in two different cars to a stadium in a distant city. At noon, Brayde

USPshnik [31]

Answer:

B(t) = -30t + 280

A(t) = -40t + 340

(Screenshot of graph is attached below)

They will be the same distance away from the stadium after 6 hours.

Step-by-step explanation:

4 0

3 years ago

What are the possible numbers of positive, negative, and complex zeros of f(x) = x6 − x5 − x4 + 4x3 − 12x2 + 12?

FrozenT [24]

Using <span>Descartes' rule of signs.
There are 4 changes of signs of coefficients:
+ to - (+x^6-x^5)
- to + (-x^4+4x^3)
+ to - (+4x^3-12x^2)
- to + (-12x^2+12)
Therefore, there are 4,2 or 0 positive roots </span><span><span>at most</span>. So it must be the third answer.
</span>

3 0

3 years ago

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g The average midterm score of students in a certain course is 70 points. From the past experience it is known that the midterm

spayn [35]

Answer:

0.98 = 98% probability that the average midterm score of these students is at most 75 points.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .