A mouse is traveling at a rate of 100 feet per hour. One half hour later, a snake smells the

Mathematics

1 answer:

Arlecino [84]3 years ago

5 0

Let, snake catches the mouse in t hours. Till then mouse is travelling from

( t + 1.5 ) hours.

Distance, travelled by mouse in t hours is, $d = 100(t+ 1.5)$ .

To catch the mouse snake had to travel the same distance d in t hours.

So,

$200t = 100( t + 1.5 )\\\\2t = t + 1.5\\\\t = 1.5 \ hours$

Therefore, snake will catch mouse in 1.5 hours.

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Combine Like Terms and Eliminate Parenthesis.<br>5z-[2z-(13z+8)]

katen-ka-za [31]

We need to start from the innermost parenthesis and work our way out.

The first parenthesis is $13z+8$ . These are not like terms because one involves a variable, while the other is a constant term. Two terms are summable if they involve the same power(s) of the same variable(s).

So, we can take one step outwards, and we arrive to the square brackets. We have

$2z-13z-8$

and 2z and -13z are like terms, so we can sum them:

$2z-(13z +8) = 2z-13z-8 = (2-13)z-8 = -11z-8$

Finally we arrive to the whole expression, which is

$5z-(-11z-8) = 5z+11z+8 = (5+11)z+8=16z+8$

Because, again, 5z and 11z were like terms.

6 0

3 years ago

. Exam scores for a large introductory statistics class follow an approximate normal distribution with an average score of 56 an

Andru [333]

Answer:

0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 56, \sigma = 5, n = 20, s = \frac{5}{\sqrt{20}} = 1.12$