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3 years ago

3. Find the perimeter of the diagram. Round to the nearest hundredth.

Mathematics

1 answer:

SpyIntel [72]3 years ago

7 0

Answer:

I think the perimeter of the diagram is 5 and if you round it like

If it's 5 or greater, add 1 to the digit in the tenths place, and then remove all the digits to the right. (In the example above, the hundredths digit is a 4 , so you would get 51.0 .) To round a number to the nearest hundredth , look at the next place value to the right (the thousandths this time).

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How many hours did jacoboys drive on Monday and Tuesday and in all?

melomori [17]

Looks like 10 and and a half if it's four it's 10 and a half if it 6 it's 12 and a half

7 0

3 years ago

The mean points obtained in an aptitude examination is 159 points with a standard deviation of 13 points. What is the probabilit

Korolek [52]

Answer:

0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled

Step-by-step explanation:

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

Normal probability distribution:

Problems of normally distributed samples are 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.

Central limit theorem:

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 $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 159, \sigma = 13, n = 60, s = \frac{13}{\sqrt{60}} = 1.68$

What is the probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled?

This is the pvalue of Z when X = 159+1 = 160 subtracted by the pvalue of Z when X = 159-1 = 158. So

X = 160

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

By the Central Limit Theorem

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

$Z = \frac{160 - 159}{1.68}$

$Z = 0.6$

$Z = 0.6$ has a pvalue of 0.7257

X = 150

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

$Z = \frac{158 - 159}{1.68}$

$Z = -0.6$

$Z = -0.6$ has a pvalue of 0.2743

0.7257 - 0.2743 = 0.4514

0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled

7 0

3 years ago

Find the sum of the following ap .1) 1/15, 1/12, 1/10,....,to 11 terms

sertanlavr [38]

Answer:

33/20

Step-by-step explanation:

1/12 - 1/15 = 5/60 - 4/60 = 1/60

d = 1/60

a_n = a_1 + d(n - 1)

a_11 = 1/15 + (1/60)(11 - 1)

a_11 = 1/15 + 1/6

a_11 = 4/60 + 10/60

a_11 = 14/60

a_11 = 7/30

a_12 = 14/60 + 1/60

a_12 = 15/60

a_12 = 1/4

s_n = n(a_1 + a_n)/2

s_11 = 11(1/15 + 7/30)/2

s_11 = 11(2/30 + 7/30)/2

s_11 = 11(9/30)/2

s_11 = 99/60

s_11 = 33/20

8 0

3 years ago

You can compare irrational numbers using rational approximation.

Natasha_Volkova [10]

Yes you can!!!!!!!!!!!

3 0

2 years ago

Can a math god help me out?

Taya2010 [7]

Answer:

$f(1)=70$

$f(n)=f(n-1)+6$

Step-by-step explanation:

One is given the following function:

$f(n)=64+6n$

One is asked to evaluate the function for $(f(1))$ , substitute $(1)$ in place of $(n)$ , and simplify to evaluate:

$f(1)=64+6(1)$

$f(1)=64+6$

$f(1)=70$

A recursive formula is another method used to represent the formula of a sequence such that each term is expressed as a function of the last term in the sequence. In this case, one is asked to find the recursive formula of an arithmetic sequence: that is, a sequence of numbers where the difference between any two consecutive terms is constant. The following general formula is used to represent the recursive formula of an arithmetic sequence:

$a_n=a_(_n_-_1_)+d$

Where ( $a_n$ ) is the evaluator term ( $a_(_n_-_1_)$ ) represents the term before the evaluator term, and (d) represents the common difference (the result attained from subtracting two consecutive terms). In this case (and in the case for most arithmetic sequences), the common difference can be found in the standard formula of the function. It is the coefficient of the variable (n) or the input variable. Substitute this into the recursive formula, then rewrite the recursive formula such that it suits the needs of the given problem,

$a_n=a_(_n_-_1_)+d$

$a_n=a_(_n_-_1_)+6$

$f(n)=f(n-1)+6$

3 0

2 years ago