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Leokris [45]
3 years ago
7

When the polynomial is written in standard form, what are the values of the leading coefficient and the constant?

Mathematics
1 answer:
raketka [301]3 years ago
3 0

Answer: The answer is C. The leading coefficient is -3 and the constants is 2.

explanation: I just took the test

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TiliK225 [7]
It would be one thousand and two

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WILL GIVE BRAINLIEST
rosijanka [135]

Answer:

increase

Step-by-step explanation:

  • Given data is: 71, 67, 67, 17, 69, 84, 21, 87

  • Arranging in ascending order we find: 17, 21, 67, 67, 69, 71, 84, 87

  • Here N = 8 (Even number)

  • N/2 = 4 and N/2 + 1 = 5

  • 4th term = 67, 5th term = 69

  • -> median = Sum of 4th and 5th term/2 = (67 + 69)/2 = 132/2 = 66

  • Now, when one of the 67 is replaced by 71, the new data set in ascending order will be: 17, 21, 67, 69, 71, 71, 84, 87

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  • -> new median > initial median

CONCLUSION: Median will increase if the number 71 replaced one of the 67's in the set.

5 0
2 years ago
For the following hypothesis test, H0: μ ≥ 150Ha: μ < 150​the test statistic a. must be positive. b. must be negative. c. can
Virty [35]

Answer:

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".  

And since we are conducting a lower unilateral lest  then we expected that the value for the statistic would be negative. And the correct answer would be:

b. must be negative

Step-by-step explanation:

Data given and notation      

\bar X represent the sample mean    

s represent the standard deviation for the sample      

n sample size      

\mu_o =150 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.      

We need to conduct a hypothesis in order to determine if the mean is lower than 150, the system of hypothesis would be:      

Null hypothesis:\mu \geq 150      

Alternative hypothesis:\mu < 150      

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".  

And since we are conducting a lower unilateral lest  then we expected that the value for the statistic would be negative. And the correct answer would be:

b. must be negative

7 0
3 years ago
I need the answers for these please
Luda [366]
11.p=-8, 17-8=9
12.y=11, 3*11=33+16=46
13.t=4, 4*4=16-14=2
14.x=9, -9x 9*8=82-9=72-9=62
15.z=4, 12*4=48-18=30
16.g=0, 4*0=0, 7+0=7
17.x=4, 9*4=36-24=-3
18.q=3, 18*3=48+2=50
19.c=2, 3*2=6-4.5=4.1
20.y=4, 9+4=13+4.8=17.4

8 0
3 years ago
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