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Zielflug [23.3K]
2 years ago
11

Very very very long essay on the topic of mom​

Mathematics
1 answer:
Leokris [45]2 years ago
4 0
For 5 points? Good luck getting someone to do that for you
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Classify the triangle by its side lengths.
Harrizon [31]

Answer:

this triangle is an acute triangle

3 0
3 years ago
F (x) = x^2 + 2x - 5<br> g(x) = 2x + 4<br> What is (f• g)(x)?
Vilka [71]

Answer:

Step-by-step explanation:

This is a composite function, f(g(x)). This means, working from the inside function to the outside, we will take the function g and plug it in for x in the f function. g(x) = 2x + 4. We will plug that into f(x) and evaluate f(2x+4):

f(g(x))=(2x+4)^2+2(2x+4)-5 and I imagine your teacher has you simplify completely. We FOIL and distribute to get

f(g(x))=4x^2+16x+16+4x+8-5 which, by combining like terms, give us

f(g(x))=4x^2+20x+19

5 0
3 years ago
Bella had already read 20% of the book before her teacher assigned it to the class. She is choosing between two options to finis
Lubov Fominskaja [6]

Answer:

How big is the book?

If it has more than 96 pages remaining 25% each night is quicker if it has less that 96 or less remaining 24 pages a night is faster

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Ed's new car has a value of $23,500, but it is expected to depreciate at a rate of 7.5% per year. If you were to write an expone
Delicious77 [7]

Answer:

0.925 or 92.5%

Step-by-step explanation:

y=a(1-b)^x

decay factor is 1-b <- and the part we care about

b is 7.5%=0.075

1-0.075

=0.925

I'm a little rusty on exponential rates, but I hope this helps!

3 0
3 years ago
A maker of a certain brand of low-fat cereal barsclaims that the average saturated fat content is 0.5gram. In a random sample of
finlep [7]

Answer:

t=\frac{0.475-0.5}{\frac{0.183}{\sqrt{8}}}=-0.386    

p_v =2*P(t_{(7)}  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is different from 0.5 at 5% of signficance.  So then the claim makes sense

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=0.475

The sample deviation calculated s=0.183

We need to conduct a hypothesis in order to check if the true mean is 0.5 or no, the system of hypothesis would be:  

Null hypothesis:\mu = 0.5  

Alternative hypothesis:\mu \neq 0.5  

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{0.475-0.5}{\frac{0.183}{\sqrt{8}}}=-0.386    

P-value

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

df=n-1=8-1=7  

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

p_v =2*P(t_{(7)}  

Conclusion  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is different from 0.5 at 5% of signficance.  So then the claim makes sense

4 0
3 years ago
Read 2 more answers
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