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Tatiana [17]
3 years ago
6

A wild horse runs at a rate of 8 miles for 6 hours. Let f(x) be the distance in miles, the horse travels for a given amount of t

ime, x, in hours. This situation can be modeled by a function. What is the domain of the function?
a. 0 < x < 6

b. 0 < f(x) < 6

c. 0 < x < 48

d. 0 < f(x) < 48
Mathematics
1 answer:
Bond [772]3 years ago
4 0

Answer:

SORRRY

Step-by-step explanation:

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How much fluid can a solid measuring 10 feet long, 12 feet wide, by 5 feet deep hold? do i need to multiply them all?
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Answer:

Persuming your asking for volume, 600 ft³

Step-by-step explanation:

If your asking for volume, just multiply all of them

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The germination rate is the rate at which plants begin to grow after the seed is planted.
chubhunter [2.5K]

Answer:

z=\frac{0.467 -0.9}{\sqrt{\frac{0.9(1-0.9)}{15}}}=-5.59  

p_v =P(z  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of germinated seeds is significantly lower than 0.9 or 90%

Step-by-step explanation:

Data given and notation

n=15 represent the random sample taken

X=7 represent the number of seeds germinated

\hat p=\frac{7}{15}=0.467 estimated proportion of seeds germinated

p_o=0.9 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)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion of germinated seeds is less than 0.9 or 90%.:  

Null hypothesis:p\geq 0.9  

Alternative hypothesis:p < 0.9  

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.

Calculate the statistic  

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

z=\frac{0.467 -0.9}{\sqrt{\frac{0.9(1-0.9)}{15}}}=-5.59  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about 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 assumed is \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(z  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of germinated seeds is significantly lower than 0.9 or 90%

4 0
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Round 199.532 to the nearest n tenth
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Graph f(x) = 6x^2-11+3 and<br> find the zeroes
babunello [35]

Answer:

x = ±1.155

Step-by-step explanation:

We need to graph the given function f(x) = 6x^2-11+3 and find its zeroes.

The attached figure shows the graph of the given function.

To find zeroes,

Put f(x) = 0

So,

6x^2-11+3=0\\\\x=\pm 1.155

Hence, this is the required solution.

5 0
3 years ago
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