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Allushta [10]
2 years ago
5

Last year,Mark was 46 inches tall.This year,Marks height is 3 inches less than peters height.Peter is 51 inches tall.How tall is

mark?
Write an equation to represent the situation.
Then solve the equation to answer the problem.
Mathematics
2 answers:
In-s [12.5K]2 years ago
7 0
Im not sure abbout the equations... but Mark is 48 inches, cuz it says; He WAS 46 in. tall. this year, amrks height is 3 inches LESS than peters height. So now, peter is 51 in. tall How tall is mark NOW? So u just have to minus 51-3 = 48! 
zvonat [6]2 years ago
4 0
I think the equation is 51-3=49 
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slavikrds [6]

Answer: 2.1 Pull out like factors :

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Equation at the end of step

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Pulling out like terms

4.1     Pull out like factors :  6s - 4r  =   -2 • (2r - 3s)

Equation at the end of step

 ((0--18•(2r-3s))+3s)--14•(r-3s)

Final result :

 <u>50r - 93s</u>

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2 years ago
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The weight of corn chips dispensed into a 14-ounce bag by the dispensing machine has been identified as possessing a normal dist
mamaluj [8]
The probability that a normally distributed dataset with a mean, μ, and statndard deviation, σ, exceeds a value x, is given by

P(X\ \textgreater \ x)=1-P(X\ \textless \ x)=1-P\left(z\ \textless \  \frac{x-\mu}{\frac{\sigma}{\sqrt{n}}} \right)

Given that t<span>he weight of corn chips dispensed into a 14-ounce bag by the dispensing machine is a normal distribution with a mean of 14.5 ounces and a standard deviation of 0.2 ounce.

</span>If <span>100 bags of chips are randomly selected the probability that the mean weight of these 100 bags exceeds 14.6 ounces is given by

P(X\ \textgreater \ 14.6)=1-P\left(z\ \textless \  \frac{14.6-14.5}{\frac{0.2}{\sqrt{100}}} \right) \\  \\ =1-P\left(z\ \textless \  \frac{0.1}{\frac{0.2}{10}} \right)=1-P\left(z\ \textless \  \frac{0.1}{0.02} \right) \\  \\ =1-P(z\ \textless \ 5)=1-1=0

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7 0
2 years ago
Suppose that the Department of Transportation would like to test the hypothesis that the average age of cars on the road is less
dimulka [17.4K]

Answer:

z=\frac{10.6-12}{\frac{4.1}{\sqrt{45}}}=-2.29  

z_{critc}= -1.64

Since our calculates values is lower than the critical value we have enough evidence to reject the null hypothesis at 5% of significance. And the best conclusion for this case is:

B. Since the test statistic is less than the critical value, we can conclude that the average age of cars on the road is less than 12 years.

Step-by-step explanation:

Data given and notation  

\bar X=10.6 represent the sample mean

\sigma=4.1 represent the population standard deviation

n=45 sample size  

\mu_o =12 represent the value that we want to test  

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

z 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 less than 12, the system of hypothesis would be:  

Null hypothesis:\mu \geq 12  

Alternative hypothesis:\mu < 12  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-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:  

z=\frac{10.6-12}{\frac{4.1}{\sqrt{45}}}=-2.29  

Critical value

For this case since w ehave a left tailed distribution we need to find a value who accumulates 0.05 of the area on th left in the normal standard distribution, and we can use the following excel code:

"=NORM.INV(0.05,0,1)"

z_{critc}= -1.64

Conclusion  

Since our calculates values is lower than the critical value we have enough evidence to reject the null hypothesis at 5% of significance. And the best conclusion for this case is:

B. Since the test statistic is less than the critical value, we can conclude that the average age of cars on the road is less than 12 years.

3 0
2 years ago
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