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kupik [55]
3 years ago
15

At Luby's, an adult's meal is $7.50 and a kid's meal

Mathematics
1 answer:
jasenka [17]3 years ago
8 0
Before i try and answer this do u have a graph
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I AM IN NEED OF HELP THANKYOU
Varvara68 [4.7K]
Oh okay I’m not going home I’m sorry I don’t have to sleep tueitu lol I
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3 years ago
The n candidates for a job have been ranked 1, 2, 3,…, n. Let X 5 the rank of a randomly selected candidate, so that X has pmf p
Anon25 [30]

Answer:

A. E(x) = 1/n×n(n+1)/2

B. E(x²) = 1/n

Step-by-step explanation:

The n candidates for a job have been ranked 1,2,3....n. Let x be the rank of a randomly selected candidate. Therefore, the PMF of X is given as

P(x) = {1/n, x = 1,2...n}

Therefore,

Expectation of X

E(x) = summation {xP(×)}

= summation {X×1/n}

= 1/n summation{x}

= 1/n×n(n+1)/2

= n+1/2

Thus, E(x) = 1/n×n(n+1)/2

Value of E(x²)

E(x²) = summation {x²P(×)}

= summation{x²×1/n}

= 1/n

3 0
3 years ago
Find the 64th term in the number sequence<br> 4,9,14,19
Karolina [17]

Answer:

319 is the term

Step-by-step explanation:

Please give me brainliest :)

8 0
2 years ago
Read 2 more answers
If 12,025 was invested at 3.25% compounded monthly 3 years, how much interest will be earned
wariber [46]

Answer:

$1,229.75

Step-by-step explanation:

Lets use the compound interest formula provided to solve this:

A=P(1+\frac{r}{n} )^{nt}

<em>P = initial balance</em>

<em>r = interest rate (decimal)</em>

<em>n = number of times compounded annually</em>

<em>t = time</em>

<em />

First, change 3.25% into a decimal:

3.25% -> \frac{3.25}{100} -> 0.0325

Since the interest is compounded monthly, we will use 12 for n. Lets plug in the values now:

A=12,025(1+\frac{0.0325}{12})^{12(3)}

A=13,254.75

Lastly, subtract A from P to get the interest earned:

13,254.75 - 12,025 = 1,229.75

6 0
3 years ago
In a recent Super Bowl, a TV network predicted that 50 % of the audience would express an interest in seeing one of its forthcom
coldgirl [10]

Answer:

z= -0.968

We can conclude that we fail to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who says that  they would watch one of the television shows not differs from 0.5 or 50% .  

Step-by-step explanation:

1) Data given and notation n  

n=106 represent the random sample taken

X=48 represent the people who says that  they would watch one of the television shows.

\hat p=\frac{48}{106}=0.453 estimated proportion of people who says that  they would watch one of the television shows.

p_o=0.5 is the value that we want to test

\alpha represent the significance level  

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that 50% of people who says that  they would watch one of the television shows.:  

Null hypothesis:p=0.5  

Alternative hypothesis:p \neq 0.5  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.453 -0.5}{\sqrt{\frac{0.5(1-0.5)}{106}}}=-0.968  

4) Statistical decision  

P value method or p value approach . "This method consists on determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level is not provided, but we can assume \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

p_v =2*P(z  

So based on the p value obtained and using the significance level assumed \alpha=0.05 we have p_v>\alpha so we can conclude that we fail to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who says that  they would watch one of the television shows not differs from 0.5 or 50% .  

6 0
3 years ago
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