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EastWind [94]
3 years ago
7

List the fractors of 35 is this number prime or composite

Mathematics
2 answers:
Nesterboy [21]3 years ago
4 0

Answer: Composite

Step-by-step explanation: To see if 35 is a prime number or a composite number, let's first start by finding the factors of 35.

To find the factors of 35, begin by dividing 35 by 1 which gives us 35. This means that 1 and 35 are factors.

Next, since 35 is not divisible by 2, 3, or 4, we divided 35 by 5 which gives us 7. This means that 5 and 7 are factors.

Next, since 35 is not divisible by 6, we divide 35 by 7 which gives us 5. Notice that 7 and 5 are repeat factors and once we hit repeat factors, this means that all other factors we try will repeat.

Therefore, the factors of 35 are 1, 5, 7, and 35.

Since 35 has more than two factors, it's called a composite number.

kenny6666 [7]3 years ago
3 0
35 is a composite number. 35=5×7. 
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Consider a chemical company that wishes to determine whether a new catalyst, catalyst XA-100, changes the mean hourly yield of i
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Answer:

Null hypothesis:\mu = 750  

Alternative hypothesis:\mu \neq 750  

t=\frac{811-750}{\frac{19.647}{\sqrt{5}}}=6.943  

p_v =2*P(t_{4}>6.943)=0.00226  

If we compare the p value and a significance level assumed \alpha=0.05 we see that p_v so we can conclude that we reject the null hypothesis, and the actual true mean is significantly different from 750 pounds per hour.  

Step-by-step explanation:

Data given and notation

Data:    801, 814, 784, 836,820

We can calculate the sample mean and sample deviation with the following formulas:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X=811 represent the sample mean  

s=19.647 represent the standard deviation for the sample

n=5 sample size  

\mu_o =750 represent the value that we want to test  

\alpha 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 to be tested  

We need to conduct a hypothesis in order to determine if the mean is different from 750 pounds per hour, the system of hypothesis would be:  

Null hypothesis:\mu = 750  

Alternative hypothesis:\mu \neq 750  

Compute the test statistic  

We don't know the population deviation, so for this case 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".  

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

t=\frac{811-750}{\frac{19.647}{\sqrt{5}}}=6.943  

Now we need to find the degrees of freedom for the t distirbution given by:

df=n-1=5-1=4

What do you conclude?  

Compute the p-value  

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

p_v =2*P(t_{4}>6.943)=0.00226  

If we compare the p value and a significance level assumed \alpha=0.05 we see that p_v so we can conclude that we reject the null hypothesis, and the actual true mean is significantly different from 750 pounds per hour.  

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I paid $42 for six hot chocolate bombs what is the unit rate price for one item 
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Answer:

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Step-by-step explanation:

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