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Stella [2.4K]
3 years ago
8

The attendance at a family reunion was 160 people. This was 125% of last year's attendance. How many people attended the reunion

last year.
Mathematics
1 answer:
Alex787 [66]3 years ago
4 0
If you would like to know how many people attended the reunion last year, you can calculate this using the following steps:

125% of last year's attendance is 160 people
125% of x is 160
125% * x = 160
125/100 * x = 160     /*100/125
x = 160 * 100 / 125
x = 128 people (last year)

Result: 128 people attended the reunion last year.
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An industrial expert claims that the average useful lifetime of a typical car transimssion which comes with ten years warranty i
lesya692 [45]

Answer:

a. none of these answers

Data provides sufficient evidence, at 1% significance level, to support the expert's claim. In addition the p-value (or the observed significance level) is equal to P( T >3.281)

Step-by-step explanation:

Data given and notation  

\bar X=13.5 represent the mean height for the sample  

s=3.2 represent the sample standard deviation for the sample  

n=9 sample size  

\mu_o =10 represent the value that we want to test

\alpha=0.01 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)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 10 years, the system of hypothesis would be:  

Null hypothesis:\mu \leq 10  

Alternative hypothesis:\mu > 10  

If we analyze the size for the sample is < 30 and 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=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{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".  

Calculate the statistic

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

t=\frac{13.5-10}{\frac{3.2}{\sqrt{9}}}=3.28    

P-value

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

df=n-1=9-1=8  

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

p_v =P(t_{(8)}>3.281)=0.00558  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can't conclude that we have a mean higher than 10 years at 1% of significance.  

a. none of these answers

3 0
3 years ago
Can you help me find the code<br><br><br><br> link/picture in the comments
Margaret [11]

Answer:

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Step-by-step explanation:

sussy baka????

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5 0
3 years ago
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4(px + 1) = 64<br> what is the value of x in terms of p ?
Sedaia [141]

Answer:

x = 15/p

Step-by-step explanation:

4(px + 1) = 64

4px + 4 = 64

-4 -4

4px/p = 60/p

4x/4 = 60/p/4

x = 15/p

4 0
3 years ago
The University of Washington claims that it graduates 85% of its basketball players. An NCAA investigation about the graduation
Nonamiya [84]

Probabilities are used to determine the chances of events

The given parameters are:

  • Sample size: n = 20
  • Proportion: p = 85%

<h3>(a) What is the probability that 11 out of the 20 would graduate? </h3>

Using the binomial probability formula, we have:

P(X = x) = ^nC_x p^x(1 - p)^{n -x}

So, the equation becomes

P(x = 11) = ^{20}C_{11} \times (85\%)^{11} \times (1 - 85\%)^{20 -11}    

This gives

P(x = 11) = 167960 \times (0.85)^{11} \times 0.15^{9}

P(x = 11) = 0.0011

Express as percentage

P(x = 11) = 0.11\%

Hence, the probability that 11 out of the 20 would graduate is 0.11%

<h3>(b) To what extent do you think the university’s claim is true?</h3>

The probability 0.11% is less than 50%.

Hence, the extent that the university’s claim is true is very low

<h3>(c) What is the probability that all  20 would graduate? </h3>

Using the binomial probability formula, we have:

P(X = x) = ^nC_x p^x(1 - p)^{n -x}

So, the equation becomes

P(x = 20) = ^{20}C_{20} \times (85\%)^{20} \times (1 - 85\%)^{20 -20}    

This gives

P(x = 20) = 1 \times (0.85)^{20} \times (0.15\%)^0

P(x = 20) = 0.0388

Express as percentage

P(x = 20) = 3.88\%

Hence, the probability that all 20 would graduate is 3.88%

<h3>(d) The mean and the standard deviation</h3>

The mean is calculated as:

\mu = np

So, we have:

\mu = 20 \times 85\%

\mu = 17

The standard deviation is calculated as:

\sigma = np(1 - p)

So, we have:

\sigma = 20 \times 85\% \times (1 - 85\%)

\sigma = 20 \times 0.85 \times 0.15

\sigma = 2.55

Hence, the mean and the standard deviation are 17 and 2.55, respectively.

Read more about probabilities at:

brainly.com/question/15246027

8 0
3 years ago
Fina bought a DVD player for $58.97 and a set of DVDs for $24.98. How much did Fina spend in all?
Softa [21]
Fina spent $83.65 in all.
4 0
2 years ago
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