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Shkiper50 [21]
3 years ago
15

Please help explain the steps

Mathematics
1 answer:
Gennadij [26K]3 years ago
3 0

Step-by-step explanation:

3a^2+24a+45=0

×3/3( first ×3)

9a^2+24(3a)+135=0

(3a+15)(3a+9)=0

now /3

3(a+5)(a+3)

a=-5✓

a=-3✓

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Is 3/2 greater than 1? Or Less than
Simora [160]

Answer:

3/2 is greater than 1

Step-by-step explanation:

A little trick to immediately know if a number is more or less than when it involves fractions versus whole numbers is that improper fractions will always be greater than 1.

If you want to confirm this, dividing 3/2 equals 1.5, ergo 1.5 > 1.

In essence, improper fractions will always represent a number greater than one contrasted to proper fractions because if you put it on a perspective (let's say you have 13 pie slices but you only needed 12, that means there is a pie slice that's extra that makes it greater than the original pie slices).

Hope that helps :)

6 0
3 years ago
A mathematics journal has accepted 15 articles for publication. However, due to budgetary restraints only 9 articles can be publ
Katena32 [7]
15!/(15-9)!= 32432400

i hate these so much hope it helps! :)
4 0
4 years ago
When the smaller of two consecutive integers is added to three times the larger the result is 31. find the integer
wariber [46]
8 and 10 are the integers
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3 years ago
At a football stadium, 4% of the fans in attendance were teenagers. If there were 120 teenagers at the football stadium, what wa
Law Incorporation [45]
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3 0
3 years ago
Daily-high temperature measurements for 40 consecutive days are recorded for a particular city. The mean daily-high temperature
Vesnalui [34]

Answer:

t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108    

p_v =P(t_{39}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.  

(D) P-val=0.021, fail to reject the null hypothesis

Step-by-step explanation:

1) Data given and notation  

\bar X=21.5 represent the mean for the temperatures

s=1.5 represent the sample standard deviation

n=40 sample size  

\mu_o =22 represent the value that we want to test

\alpha=0.01 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 is less than 22C, the system of hypothesis would be:  

Null hypothesis:\mu \geq 22  

Alternative hypothesis:\mu < 22  

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{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108    

P-value

We can calculate the degrees of freedom like this:

df=n-1=40-1=39

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

p_v =P(t_{39}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.  

The best option would be:

(D) P-val=0.021, fail to reject the null hypothesis

5 0
3 years ago
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