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3 years ago

The reciprocal of 2 1/7 is what ?

Mathematics

2 answers:

aivan3 [116]3 years ago

6 0

Answer:

7/15.

Step-by-step explanation:

2 1/7 = 15/7

Reciprocal of 15/7 = 1 / 15/7

= 7/15.

Phantasy [73]3 years ago

5 0

Answer:

7/15

Step-by-step explanation:

2 1/7 is mixed fraction, if you turn it into a improper fraction, you will get 15/7. Just flip those two numbers, and you get the answer

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20 POINTS HELPPPPP

kiruha [24]

For this case we have the following expression:

$\frac {216 ^ {n-2}} {(\frac {1} {36}) ^ {3n}} = 216$

We multiply both sides by: $(\frac {1} {36}) ^ {3n}$

$216 ^ {n-2} = 216 * (\frac {1} {36}) ^ {3n}$

We divide both sides by 216:

$\frac {216 ^ {n-2}} {216} = (\frac {1} {36}) ^ {3n}$

To divide powers of the same base, we place the same base and subtract the exponents:

$216 ^ {n-2-1} = (\frac {1} {36}) ^ {3n}\\216 ^ {n-3} = (\frac {1} {36}) ^ {3n}$

Rewriting:

$(6 ^ 3) ^ {n-3} = (\frac {1} {6 ^ 2}) ^ {3n}\\6 ^ {3n-9} = \frac {1} {6 ^ {6n}}\\6^{ 3n-9} * 6^{ 6n} = 1$

To multiply powers of the same base, we place the same base and add the exponents:

$6^{ 3n-9 + 6n} = 1\\6^{ 9n-9} = 1$

We know that any number raised to zero is 1, $a ^ 0 = 1.$

So, for equality to be true:

$9n-9 = 0\\9n = 9\\n = \frac {9} {9}\\n = 1$

Answer:

$n = 1$

3 0

3 years ago

Read 2 more answers

Two friends are collecting stuffed animals. Veronica has 5 more than double the number of stutted

STatiana [176]

v + m = 32 and v = 5 + 2m are the equations that are used to determine m, the number of stuffed animals Mariposa has

Number of stuffed animals Mariposa has is 9 and number of stuffed animals with Veronica is 23

<h3><u>Solution:</u></h3>

Let "v" be the number of stuffed animals with Veronica

Let "m" be the number of stuffed animals with Mariposa

Given that,

Together, they have 32 stuffed animals

Therefore,

v + m = 32 --------- eqn 1

Veronica has 5 more than double the number of stutted animals as her friend Mariposa

Therefore,

Number of stuffed animals with Veronica = 5 + 2(number of stuffed animals with Mariposa)

v = 5 + 2m ---------- eqn 2

Thus eqn 1 and eqn 2 can be used to determine m, the number of stuffed animals Mariposa has

Let us solve eqn 1 and eqn 2

Substitute eqn 2 in eqn 1

5 + 2m + m = 32

5 + 3m = 32

3m = 32 - 5

3m = 27

Substitute m = 9 in eqn 2

v = 5 + 2(9)

v = 5 + 18

Thus number of stuffed animals Mariposa has is 9 and number of stuffed animals with Veronica is 23

5 0

3 years ago

I need helpppp !!! :((

Andrew [12]

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Step-by-step explanation:

3 0

3 years ago

How do you solve a-9/a3

Klio2033 [76]

Is it this?
$\frac{a - 9}{a3}$
If so, then
$\frac{a - 9}{a3} = 0 \\ \frac{a}{a3} - \frac{9}{a3} = 0 \\ \frac{1}{3} - \frac{9}{a3} \\ \frac{1}{3} = \frac{9}{a3} \\ 3( \frac{1}{3} ) = 3( \frac{9}{a3} ) \\ 1 = \frac{9}{a} \\ a = 9$

5 0

3 years ago

A simple random sample of size nequals10 is obtained from a population with muequals68 and sigmaequals15. (a) What must be true

valentina_108 [34]

Answer:

(a) The distribution of the sample mean ( $\bar x$ ) is <em>N</em> (68, 4.74²).

(b) The value of $P(\bar X$ is 0.7642.

Step-by-step explanation:

A random sample of size <em>n</em> = 10 is selected from a population.

Let the population be made up of the random variable <em>X</em>.

The mean and standard deviation of <em>X</em> are:

$\mu=68\\\sigma=15$

(a)

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Since the sample selected is not large, i.e. <em>n</em> = 10 < 30, for the distribution of the sample mean will be approximately normally distributed, the population from which the sample is selected must be normally distributed.

Then, the mean of the distribution of the sample mean is given by,

$\mu_{\bar x}=\mu=68$