Anna said that the product of 7/8 ⋅ 1 1/2 =7/2. How can you tell that her answer is wrong/

Answers

Answer 1

Answer:

Answer:

The answer is 77/16, but you can go much dipper if you want. By just simplify 77/16, in 4×13/16. And the final result will be 4.8125. Please if you find it helpful give me a like and rate me.

Answer 2

Answer:

Answer:

it's 77/16 and just say Anna's answer is wrong.

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Judging on the basis of experience, a politician claims that 57% of voters in a certain area have voted for an independent candidate in past elections. Suppose you surveyed 25 randomly selected people in that area, and 18 of them reported having voted for an independent candidate. The null hypothesis is that the overall proportion of voters in the area that have voted for an independent candidate is 57%. What value of the test statistic should you report?
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You drop a rock off a bridge. The rock’s height, h (in feet above the water), after t seconds is modeled by h = -16t^2 + 541. What is the height of the rock after 2 seconds?

I feel nice today, Here is 100 points

Answers

Answer:

oh well thanks

Step-by-step explanation:

Thank you sooo much l!!!!!

Please help me I beg!

Answers

Answer:

38.9

Step-by-step explanation:

Giving a test to a group of students, the grades and gender are summarized below. Round your answers to 4 decimal places. A B C Total
Male 16 13 6 35
Female 2 3 8 13
Total 18 16 14 48
If one student is chosen at random:
1. Find the probability that the student was female AND got a "B".
2. Find the probability that the student was male AND got a "A".
3. Find the probability that the student got a B.

Answers

Answer:

1). 0.1667

2). 0.3333

3). 0.3333

Step-by-step explanation:

1). Probability that the student was female and got a 'B'

= $\frac{\text{Number of female students who got B}}{\text{Total number of students}}$

= $(3)/(48)=(1)/(16)$

= 0.1667

2). Probability that the student was male and got an 'A'

= $\frac{\text{Number of male students who got A }}{\text{Total number of students}}$

= $(16)/(48)$

= $(1)/(3)$

= 0.3333

3). Probability that the student got a B = $\frac{\text{Number of students who got B}}{\text{Total number of students}}$

= $(16)/(48)$

= $(1)/(3)$

= 0.3333

A cell site is a site where electronic communications equipment is placed in a cellular network for the use of mobile phones. The numbers y of cell sites from 1985 through 2011 can be modeled byy = 269573/1+985e^-0.308t where t represents the year, with t = 5 corresponding to 1985. Use the model to find the numbers of cell sites in the years 1998, 2003, and 2006.

Answers

Answer:

(a) 3178

(b) 14231

Step-by-step explanation:

Given

$y = (269573)/(1+985e^(-0.308t))$

Solving (a): Year = 1998

1998 means t = 8 i.e. 1998 - 1990

So:

$y = (269573)/(1+985e^(-0.308*8))$

$y = (269573)/(1+985e^(-2.464))$

$y = (269573)/(1+985*0.08509)$

$y = (269573)/(84.81365)$

$y = 3178$ --- approximated

Solving (b): Year = 2003

2003 means t = 13 i.e. 2003 - 1990

So:

$y = (269573)/(1+985e^(-0.308*13))$

$y = (269573)/(1+985e^(-4.004))$

$y = (269573)/(1+985*0.01824)$

$y = (269573)/(18.9664)$

$y = 14213$ --- approximated

Solving (c): Year = 2006

2006 means t = 16 i.e. 2006 - 1990

So:

$y = (269573)/(1+985e^(-0.308*16))$

$y = (269573)/(1+985e^(-4.928))$

$y = (269573)/(1+985*0.00724)$

$y = (269573)/(8.1314)$

$y = 33152$ --- approximated

A digital camcorder repair service has set a goal not to exceed an average of 5 working days from the time the unit is brought in to the time repairs are completed. A random sample of 12 repair records showed the following repair times (in days): 5, 7, 4, 6, 7, 5, 5, 6, 4, 4, 7, 5.(a) H0: μ ≤ 5 days versus H1: μ > 5 days. At α = .05, choose the right option. Reject H0 if tcalc > 1.7960
Reject H0 if tcalc < 1.7960

b. Calculate the Test statistic.

c-1. The null hypothesis should be rejected.
i. TRUE
ii. FALSE

c-2. The average repair time is longer than 5 days.
i. TRUE
ii. FALSE

c-3 At α = .05 is the goal being met?
i. TRUE
ii. FALSE

Answers

Answer:

a) Reject H0 if tcalc > 1.7960

b) $t=(5.42-5)/((1.16)/(√(12)))=1.239$

c-1) ii. FALSE

c-2) ii.FALSE

c-3)i. TRUE

Step-by-step explanation:

1) Data given and notation

$\bar X=5.42$ represent the mean time for the sample

$s=1.16$ represent the sample standard deviation for the sample

$n=12$ sample size

$\mu_o =5$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

a) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 5 days, the system of hypothesis would be:

Null hypothesis: $\mu \leq 5$

Alternative hypothesis: $\mu > 5$

We don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=(\bar X-\mu_o)/((s)/(√(n)))$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Rejection zone

On this case we need a critical value that accumulates 0.05 of the area on the right tail. The degrees of freedom are given by 11. And we can use the following excel code to find the critical value : "T.INV(1-0.95,11)" and the critical value would be given by $t_(\alpha/2)=1.795$ .

And the rejection zone is given by:

Reject H0 if tcalc > 1.7960

b) Calculate the statistic

We can replace in formula (1) the info given like this:

$t=(5.42-5)/((1.16)/(√(12)))=1.239$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=12-1=11$

Since is a one side test the p value would be:

$p_v =P(t_((11))>1.239)=0.121$

c-1. The null hypothesis should be rejected.

ii. FALSE

c-2. The average repair time is longer than 5 days.

ii. FALSE

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, and the true mean is not significantly higher than 5.

c-3 At α = .05 is the goal being met?

i. TRUE

We fail to reject the null hypothesis so then the goal is met.

An office will increase salary to its top 8% employees on the basis of a performance score the office created for each employee. The performance score is approximately normal with mean 82.5 and standard deviation 9.25. How high must an employee score in order to qualify for increase in the salary?

Answers

Answer:

An employee's score in order to qualify for increase in the salary must be higher than 95.45.

Step-by-step explanation:

Let X represent the performance score of employees.

It is provided that X follows a normal distribution with parameters μ = 82.5 and σ - 9.25.

It is provided that the office will increase salary of its top 8% employees on the basis of a performance score the office created for each employee.

That is, the probability to qualify for increase in the salary is,

P (X > x) = 0.08

⇒ P (X < x) = 0.92

⇒ P (Z < z) = 0.92

The corresponding z-value is,

z = 1.40

Compute the value of x as follows:

$z=(x-\mu)/(\sigma)\n\n1.40=(x-82.5)/(9.25)\n\nx=82.5+(1.40* 9.25)\n\nx=95.45$

Thus, an employee's score in order to qualify for increase in the salary must be higher than 95.45.

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