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Y_Kistochka [10]
3 years ago
10

One beaver can chew though a tree that is 15 cm thick in 20 min if it works at the same pase how long will it tack one beaver to

cut down enough trees to create a stack 1.2 meters tall
Mathematics
1 answer:
Vesna [10]3 years ago
4 0
It takes 8 trees to make a stack 1.2 metres tall, so if ut takes 20 minutes per tree, then it will take 8 × 20 or 160 minutes, or 2 hours 40 minutes.
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fiasKO [112]
B. x > -4 I did da test. That's da correct one :-)
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3 years ago
A researcher claims that the mean of the salaries of elementary school teachers is greater than the mean of the salaries of seco
Brut [27]

Answer:

t=\frac{(48250-45630)}{\sqrt{\frac{3900^2}{26}+\frac{5530^2}{24}}}}=1.921  

df=n_{A}+n_{B}-2=26+24-2=48

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

p_v =P(t_{(48)}>1.921)=0.0303

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the mean for elementary school teachers is significantly higher than the mean for secondary teachers at 5% of significance

Step-by-step explanation:

Data given and notation

\bar X_{A}=48250 represent the mean of elementary teachers

\bar X_{B}=45630 represent the mean for secondary teachers

s_{A}=3900 represent the sample standard deviation for elementary teacher

s_{B}=5530 represent the sample standard deviation for secondary teachers

n_{A}=26 sample size selected

n_{B}=24 sample size selected  

\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)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean of the salaries of elementary school teachers is greater than the mean of the salaries of secondary school teachers, the system of hypothesis would be:

Null hypothesis:\mu_{A}-\mu_{B}\leq 0

Alternative hypothesis:\mu_{A}-\mu_{B}>0

We don't know the population deviations, so for this case is better apply a t test to compare means, and the statistic is given by:

t=\frac{(\bar X_{A}-\bar X_{B})}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}} (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

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

t=\frac{(48250-45630)}{\sqrt{\frac{3900^2}{26}+\frac{5530^2}{24}}}}=1.921  

P-value

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

df=n_{A}+n_{B}-2=26+24-2=48

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

p_v =P(t_{(48)}>1.921)=0.0303

Conclusion

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the mean for elementary school teachers is significantly higher than the mean for secondary teachers at 5% of significance

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Aleksandr [31]
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6 0
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ASHA 777 [7]
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h= \frac{-b}{2a}
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k=128-200
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So the vertex of our function is (8,-72)

Next, we are going to use the vertex to rewrite our quadratic equation:
y=a(x-h)^2+k
y=2(x-8)^2+(-72)
y=2(x-8)^2-72
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We can conclude that:
The rewritten equation is y=2(x-8)^2-72
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Answer:

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