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3 years ago

Encuentra el cociente ??

Mathematics

1 answer:

vagabundo [1.1K]3 years ago

4 0

For this case we must find the quotient of the following expression:

$\frac {(- 8x ^ 2 * y) (- 7x ^ 5 * y ^ 3)} {- 2x ^ 2 * y ^ 3}$

We can rewrite the expression as:

(We take into account that $- * - = +$ )

$\frac {56x ^ 2 * y * x ^ 5 * y ^ 3} {- 2x ^ 2 * y ^ 3} =$

By definition of multiplication of powers of the same base, we place the same base and add the exponents:

$\frac {56x ^ {2 + 5} * y ^ {1 + 3}} {- 2x ^ 2 * y ^ 3} =\\\frac {56x ^ {7} * y ^ {4}} {- 2x ^ 2 * y ^ 3} =$

By definition of division of powers of the same base, we place the same base and subtract the exponents:

$-28x^{ 7-2} * y^{ 4-3} =\\-28x ^ 5 * y$

Answer:

$-28x ^ 5 * y$

You might be interested in

Location is known to affect the number, of a particular item, sold by an automobile dealer. Two different locations, A and B, ar

yKpoI14uk [10]

Answer:

We conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

Step-by-step explanation:

We are given that Location A was observed for 18 days and location B was observed for 13 days.

On average, location A sold 39 of these items with a sample standard deviation of 8 and location B sold 49 of these items with a sample standard deviation of 4.

Let $\mu_1$ = true mean number of sales at location A.

$\mu_2$ = true mean number of sales at location B

So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2\geq0$ or $\mu_1 \geq \mu_2$ {means that the true mean number of sales at location A is greater than or equal to the true mean number of sales at location B}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2$ or $\mu_1< \mu_2$ {means that the true mean number of sales at location A is fewer than the true mean number of sales at location B}

The test statistics that would be used here Two-sample t test statistics as we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n__1_-_n__2-2$

where, $\bar X_1$ = sample average of items sold at location A = 39

$\bar X_2$ = sample average of items sold at location B = 49

$s_1$ = sample standard deviation of items sold at location A = 8

$s_2$ = sample standard deviation of items sold at location B = 4

$n_1$ = sample of days location A was observed = 18

$n_2$ = sample of days location B was observed = 13

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(18-1)\times 8^{2}+(13-1)\times 4^{2} }{18+13-2} }$ = 6.64

So, test statistics = $\frac{(39-49)-(0)}{6.64 \times \sqrt{\frac{1}{18}+\frac{1}{13} } }$ ~ $t_2_9$

= -4.14

The value of t test statistics is -4.14.

Now, at 0.01 significance level the t table gives critical value of -2.462 at 29 degree of freedom for left-tailed test.

Since our test statistics is less than the critical values of t as -2.462 > -4.14, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

3 0

3 years ago

Ayuda porfa, quien entiende???

aksik [14]

Answer:

Step-by-step explanation:

the right answer is c it good

6 0

2 years ago

An electronics store sells about 40 MP3 players per month for $90 each. For each $5 decrease in price, the store expects to sell

erma4kov [3.2K]

Step-by-step explanation:

Let's establish our equation first:

for every $5 decrease, there's an additional of 4 MP3 players sold.
to get the monthly revenue, we need to multiple the cost of each player to the number of units sold

$\frac{40mp3}{month} \times \frac{90dollars}{mp3} = 3600 \frac{dollars}{month}$