Actually help pls people have been typing random stuff

Katen [24]

Last answer is right

3 0

3 years ago

A variety of stores offer loyalty programs. Participating shoppers swipe a bar-coded tag at the register when checking out and r

dsp73

Answer:

$t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236$

Now we can find the degrees of freedom:

$df=n-1=80-1=79$

Now we can calculate the p value with the alternative hypothesis using the following probability:

$p_v =P(t_{(79)}>2.236)=0.0141$

Using a significance level of 0.01 or 1% we have enough evidence to FAIL to reject the null hypothesis and we can conclude that shoppers participating in the loyalty program NOT spent more on average than typical shopper at this significance level assumed

Step-by-step explanation:

Information given

$\bar X=130$ represent the sample mean

$s=40$ represent the sample standard deviation

$n=80$ sample size

$\mu_o =120$ represent the value to verify

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to verify if shoppers participating in the loyalty program spent more on average than typical shoppers (120) , the system of hypothesis would be:

Null hypothesis: $\mu \leq 120$

Alternative hypothesis: $\mu > 120$

The statistic for this case is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236$

Now we can find the degrees of freedom:

$df=n-1=80-1=79$

Now we can calculate the p value with the alternative hypothesis using the following probability:

$p_v =P(t_{(79)}>2.236)=0.0141$

Using a significance level of 0.01 or 1% we have enough evidence to FAIL to reject the null hypothesis and we can conclude that shoppers participating in the loyalty program NOT spent more on average than typical shopper at this significance level assumed

3 0

4 years ago

There are 7 Democrats and 6 Republicans on a local city council. A planning committee of 6 for affordable housing needs to be

Brut [27]

Answer:

The probability is $0.383$ .

Step-by-step explanation:

There are 7 democrats and 6 republicans

Number of ways in which 4 democrats and 2 republicans can be selected= ${}^{7}{{C}_{4}}\times {}^{6}{{C}_{2}}$

Number of ways in which 5 democrats and 1 republican can be selected= ${}^{7}{{C}_{5}}\times {}^{6}{{C}_{1}}$

Number of ways in which 6 democrats can be selected = ${}^{7}{{C}_{6}}$

So, favourable outcomes are ${}^{7}{{C}_{4}}\times {}^{6}{{C}_{2}}+{}^{7}{{C}_{5}}\times {}^{6}{{C}_{1}}+{}^{7}{{C}_{6}}$

Total possible outcomes = ${}^{13}{{C}_{6}}$

Now, probability= $\frac{Favourable}{Possible}=\frac{{}^{7}{{C}_{4}}\times {}^{6}{{C}_{2}}+{}^{7}{{C}_{5}}\times {}^{6}{{C}_{1}}+{}^{7}{{C}_{6}}}{{}^{13}{{C}_{6}}}=0.383$

Learn more about probability:

brainly.com/question/2561151?referrer=searchResults

7 0

3 years ago

Me podrían ayudar y explicarme el proceso please

kakasveta [241]

Answer: Encontrarás las respuestas debajo de cada explicación. Espero que consideres darme brainliest y 5 estrellas, y más importante, que hayas entendido.

Step-by-step explanation:

Esto es la regla de 3. Conoces 3 valores y necesitas encontrar un 4to valor.

Ya sabemos que 1 caja pesa 20kg,

1 - 20kg

ahora queremos saber cuantas cajas (x) equivalen a 7653kg

x - 7653kg

Hacemos un pequeño cuadro poniendo en un lado los numeros de cajas y del otro el peso.

Pongamos la cantidad de cajas en la izquierda y el peso en la derecha

$\frac{1}{x} =\frac{20kg}{7653kg}$

Procedemos a multiplicar en cruz. Solo podemos multiplicar si tenemos ambos valores. En este caso, los valores en cruz que tenemos son 1 y 7653kg porque en la otra cruz (20 y x) tenemos nuestra incognita. Y, por ultimo, dividimos entre el numero cuya cruz es con la incognita (20kg)

Esto nos deja la expresión así;

$x=\frac{1*7653kg}{20kg}$

$x=382.65$

Como una caja no puede ser un numero decimal, redondeamos.

$x=383$

Por lo tanto, se necesitan aproximadamente 383 cajas para que su peso equivalga a 7653 kg.

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La otra pregunta dice: Si una caja pesa 20kg, (1 - 20kg), ¿Cuántas cajas se necesitan para equivaler 9500kg? (x - 9500kg)

Hacemos lo mismo que arriba, ponemos la cantidad de cajas de un lado, y el peso de otro lado.

$\frac{1}{x}=\frac{20kg}{9500kg}$

Multiplicamos aquellos valores en cruz que conozcamos y dividimos por el valor cuya cruz sea con x.

$x=\frac{1*9500kg}{20kg}$

$x=475$

Esta es la cantidad de cajas que se necesitan para que su pesa equivalga a 9500kg.

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Por último, la pregunta dice, ¿Cuántos kilogramos hay en 873 cajas?

Ya sabemos que una caja pesa 20kg ( 1 - 20kg ) y ahora buscamos cuantos kg hay en 873 cajas; es decir, (873 - x)

Nos quedaría así;

$\frac{1}{873}=\frac{20kg}{x}$

Ahora, nuestra cruz es 873 y 20kg, y el divisor es 1.

$x=\frac{873*20kg}{1}$

$x=17460kg$

Este sería el peso de 873 cajas.

5 0

3 years ago

I need help it’s my final!

Natasha2012 [34]

The answer is D because linear functions represent y=mx+b and D is the one that represents it.

3 0

3 years ago