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3 years ago

The sum of reciprocals of two numbers is 5/6 and the difference is 1/6. Find the numbers.

2 answers:

8 0

Hello,
Let's assume a and b the 2 numbers with a1/a+1/a=5/6+1/6==>2/a=1==>a=2

(1)==>1/b=5/6-1/2==>1/b=5/6-3/6==>1/b=1/3==>b=3

dsp733 years ago

7 0

X=2
, y=3

this is the answer

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heights of statistics students were obtained by a teacher as part of an experiment conducted for the class. The last digit of th

hammer [34]

Answer:

1) B. The height appear to be reported because there are disproportionately more 0s and 5s.

2) A. They are likely not very accurate because they appear to be reported.

Step-by-step explanation:

The distribution table is shown below:

Last Digit Frequency

0 9

1 1

2 1

3 3

4 1

5 11

6 1

7 0

8 3

9 1

1. Based on the distribution table, we see a very disproportionate distribution. There is a high frequency of 0's and 5's. This lays credence to the heights being reported rather than measured. As such, option B is the correct answer

B. The height appear to be reported because there are disproportionately more 0s and 5s.

2. Since the heights were reported and not measured, they are most certainly not accurate. The conclusion is that the result is not accurate. As such, option A is the correct answer

A. They are likely not very accurate because they appear to be reported.

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3 years ago

5. In one year, there were 125 major new automobile types available. If you look at the miles per gallon (MPG) for these vehicle

ziro4ka [17]

Answer:

The answer is "99.82% and 86.99%".

Step-by-step explanation:

In point a:

$=P(X < 45)\\\\= P(Z < \frac{45-29.09}{5.46}) \\\\=P(Z < 2.9139)\\\\=0.9982\\\\=99.82\%$

In point b:

$=P(X >17)\\\\= P(Z > \frac{17-22.37}{4.77}) \\\\=P(Z > -1.1258)\\\\=0.8699\\\\=86.99\% \\\\$

5 0

2 years ago

Which rule can be used to find the output numbers in this table

jekas [21]

An input-output table, like the one shown below, can be used to represent a function. Each pair of numbers in the table is related by the same function rule. That rule is multiply each input number

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2 years ago

• Which expressions are equivalent to 14a + 286?

Sveta_85 [38]

Answer:

2(7a + 143)

what are the answers or are you asking us to find them

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2 years ago

What is a quick and easy way to remember explicit and recursive formulas?

Oliga [24]

I always found derivation to be helpful in remembering. Since this question is tagged as at the middle school level, I assume you've only learned about arithmetic and geometric sequences.

First, remember what these names mean. An arithmetic sequence is a sequence in which consecutive terms are increased by a fixed amount; in other words, it is an additive sequence. If $a_n$ is the $n$ th term in the sequence, then the next term $a_{n+1}$ is a fixed constant (the common difference $d$ ) added to the previous term. As a recursive formula, that's

$a_{n+1}=a_n+d$

This is the part that's probably easier for you to remember. The explicit formula is easily derived from this definition. Since $a_{n+1}=a_n+d$ , this means that $a_n=a_{n-1}+d$ , so you plug this into the recursive formula and end up with

$a_{n+1}=(a_{n-1}+d)+d=a_{n-1}+2d$

You can continue in this pattern, since every term in the sequence follows this rule:

$a_{n+1}=a_{n-1}+2d$
$a_{n+1}=(a_{n-2}+d)+2d$
$a_{n+1}=a_{n-2}+3d$
$a_{n+1}=(a_{n-3}+d)+3d$
$a_{n+1}=a_{n-3}+4d$

and so on. You start to notice a pattern: the subscript of the earlier term in the sequence (on the right side) and the coefficient of the common difference always add up to $n+1$ . You have, for example, $(n-2)+3=n+1$ in the third equation above.

Continuing this pattern, you can write the formula in terms of a known number in the sequence, typically the first one $a_1$ . In order for the pattern mentioned above to hold, you would end up with

$a_{n+1}=a_1+nd$

or, shifting the index by one so that the formula gives the $n$ th term explicitly,

$a_n=a_1+(n-1)d$

Now, geometric sequences behave similarly, but instead of changing additively, the terms of the sequence are scaled or changed multiplicatively. In other words, there is some fixed common ratio $r$ between terms that scales the next term in the sequence relative to the previous one. As a recursive formula,

$a_{n+1}=ra_n$

Well, since $a_n$ is just the term after $a_{n-1}$ scaled by $r$ , you can write

$a_{n+1}=r(ra_{n-1})=r^2a_{n-1}$

Doing this again and again, you'll see a similar pattern emerge:

$a_{n+1}=r^2a_{n-1}$
$a_{n+1}=r^2(ra_{n-2})$
$a_{n+1}=r^3a_{n-2}$
$a_{n+1}=r^3(ra_{n-3})$
$a_{n+1}=r^4a_{n-3}$

and so on. Notice that the subscript and the exponent of the common ratio both add up to $n+1$ . For instance, in the third equation, $3+(n-2)=n+1$ . Extrapolating from this, you can write the explicit rule in terms of the first number in the sequence:

$a_{n+1}=r^na_1$

or, to give the formula for $a_n$ explicitly,

$a_n=r^{n-1}a_1$

6 0

3 years ago