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1 month ago

If c (x) = StartFraction 5 Over x minus 2 EndFraction and d(x) = x + 3, what is the domain of (cd)(x)?

Mathematics

2 answers:

babunello [11.8K]1 month ago

8 0

Respuesta:

El dominio de la función (cd)(x) incluirá todos los valores reales de x, excepto x = 2.

Explicación paso a paso:

Las dos funciones son $c(x) = \frac{5}{x - 2}$ y d(x) = x + 3

Así que, (cd)(x) = $(\frac{5}{x - 2})(x + 3) = \frac{5(x + 3)}{x - 2}$

Por tanto, cuando x = 2, la función (cd)(x) no estará definida ya que una división por cero hará que (cd)(x) no tiene sentido.

Por lo tanto, el dominio de la función (cd)(x) abarcará todos los valores reales de x, excepto x = 2. (Respuesta)

Leona [12.6K]1 month ago

4 0

Respuesta:

la respuesta es b en edg2020

Explicación paso a paso:

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babunello [11817]

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Step-by-step clarification:

We know a random sample of 16 mail-order sales receipts shows a mean sale amount of $74.50 and a standard deviation of $17.25.

For internet sales, a random sample of 9 receipts gives a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal value utilized for constructing a 99% confidence interval for the true mean difference is given by;

P.Q. =

where,

= sample mean of mail-order sales = $74.50 $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ $t__n_1_+_n_2_-_2$

= sample mean of internet sales = $84.40

$\bar X_1$ = standard deviation for mail-order sales = $17.25

$\bar X_2$ = standard deviation for internet sales = $21.25

$s_1$ = number of mail-order sales receipts = 16

= number of internet sales receipts = 9 $s_2$

Furthermore,

= $n_1$ = 18.74

$n_2$

The actual mean difference between average mail-order and internet purchases is denoted by ( $s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ $\sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} }$

Thus, the 99% confidence interval for ( $\mu_1-\mu_2$ ) is expressed as;

= $\mu_1-\mu_2$ Here, the t critical value at the 0.5% significance level with 23 degrees of freedom is 2.807. =

= [$-31.82, $12.02] $(\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}$

Therefore, the 99% confidence interval for the true mean difference between average mail-order and internet purchases is [$(-31.82), $12.02].

$(74.50-84.40) \pm (2.807 \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})$

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Answer:

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lawyer [12517]

Answer:

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Step-by-step explanation:

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