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anastassius [24]

3 years ago

11

A restaurant is offering a dinner special that includes one starter and one entree. The choices are listed below. Starter: bread

sticks, soup, salad Entree: beef, fish, chicken, shrimp, pork How many possible dinner special combinations are there

2 answers:

artcher [175]3 years ago

8 0

Answer:

There are 15 combinations.

Step-by-step explanation:

A restaurant is offering a dinner special that includes one starter and one entree.

Starter: bread-sticks, soup, salad

Entree: beef, fish, chicken, shrimp, pork

So, we have 3 starters and 5 entrees.

To know the possible dinner special combinations we will simply multiply the two.

$3\times5=15$

Therefore, there are 15 combinations.

pav-90 [236]3 years ago

8 0

Answer:

15

Step-by-step explanation:

unit test on edgenuity

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A study1 conducted in July 2015 examines smartphone ownership by US adults. A random sample of 2001 people were surveyed, and th

dusya [7]

Answer:

a) Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

b) $z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{688+671}{989+1012}=0.679$

c) $z=\frac{0.696-0.663}{\sqrt{0.679(1-0.679)(\frac{1}{989}+\frac{1}{1012})}}=1.58$

d) For this case we see that $\hat p_1 > \hat p_2$ so then the answer for this cae would men

Step-by-step explanation:

Information given

$X_{1}=688$ represent the number of men with smartphone

$X_{2}=671$ represent the number of women with smartphone

$n_{1}=989$ sample of men selected

$n_{2}=1012$ sample of women selected

$p_{1}=\frac{688}{989}=0.696$ represent the proportion of men with smartphone

$p_{2}=\frac{671}{1012}=0.663$ represent the proportion of women with smartphone

$\hat p$ represent the pooled estimate of p

z would represent the statistic

$p_v$ represent the value

Part a

We want to test if we have difference in the proportion owning a smartphone between men and women, the system of hypothesis would be:

Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

Part b

The statistic for this case is given by:

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{688+671}{989+1012}=0.679$

Part c

Replacing the info given we got:

$z=\frac{0.696-0.663}{\sqrt{0.679(1-0.679)(\frac{1}{989}+\frac{1}{1012})}}=1.58$

Part d

For this case we see that $\hat p_1 > \hat p_2$ so then the answer for this cae would men

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