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yan [13]
3 years ago
9

HELP! 8th grade math!!

Mathematics
1 answer:
zhannawk [14.2K]3 years ago
5 0
The answer is c I believe
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Write an inequality to represent the situation: The temperature stayed above -15 degrees. Help!
Ivenika [448]
T > -15

Hope this helps =)
5 0
2 years ago
Read 2 more answers
Based on a sample of 100 employees a 95% confidence interval is calculated for the mean age of all employees at a large firm. Th
babymother [125]

Answer:

a

\= x  =  40.85

b

E = 5.85

Ca  

   t_c =  2.08

Cb

   t_c  =   1.282

Explanation:

From the question we are told that

     The sample size is n  =  100

     The upper limit of the 95% confidence interval is  b =  47.2 years

     The lower limit of the 95% confidence interval is   a =  34.5 years

Generally the sample mean is mathematically represented as

         \= x  =  \frac{a + b }{2}

=>    \= x = \frac{47.2 + 34.5 }{2}

=>    \= x  =  40.85

Generally the margin of error is mathematically represented as

         E =  \frac{b- a }{ 2}

=>      E =  \frac{47.2- 34.5 }{ 2}

=>      E = 5.85

Considering question C a  

From the question we are told the confidence level is  90% , hence the level of significance is    

      \alpha = (100 - 90 ) \%

=>   \alpha = 0.10

The  sample size is  n =  22

Given that the sample size is not sufficient enough i.e n < 30 we will make use of the student t distribution table  

Generally the degree of freedom is mathematically represented as

           df =  n- 1

=>        df =  22 - 1

=>        df =  21

Generally from the student  t  distribution table the critical value  of  \frac{\alpha }{2} at a degree of freedom  of  21 is  

   t_c =t_{\frac{\alpha }{2} ,  21  } =  2.08

Considering question C b

From the question we are told the confidence level is  80% , hence the level of significance is    

      \alpha = (100 - 80 ) \%

=>   \alpha = 0.20

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   t_c  =Z_{\frac{\alpha }{2} } =  1.282

4 0
2 years ago
Answer 12 and 14 please
Dahasolnce [82]
The answer to both is 4
8 0
3 years ago
Read 2 more answers
Consider a box that contains 14 red balls,12 blue balls,and 9 yellow balls.A ball is drawn at random and the color is noted and
allochka39001 [22]

Answer:

a. The probability of both balls are blue is \frac{144}{1225}.

b. The probability of getting first red and second yellow ball is \frac{18}{175}.

Step-by-step explanation:

It is given that the box contains 14 red balls,12 blue balls,and 9 yellow balls. The total number of balls is

14+12+9=35

The probability is defined as

p=\frac{\text{Favorable outcomes}}{\text{Total number of outcomes}}

Probability of getting red ball =  \frac{14}{35}

Probability of getting blue ball =  \frac{12}{35}

Probability of getting yellow ball =  \frac{9}{35}

It is given that a ball is drawn at random and the color is noted and then put back inside the box. It means first event will not effect the probability of second event.

a.

The probability of both balls are blue is

\frac{12}{35}\times \frac{12}{35}=\frac{144}{1225}

b.

The probability of getting first red and second yellow ball is

\frac{14}{35}\times \frac{9}{35}=\frac{18}{175}

6 0
3 years ago
Read 2 more answers
Please help! ;)------
AlexFokin [52]

Answer:

The answer is c

Step-by-step explanation:

When there is a division with the sae variable and different exponent, you substract the exponent from the denominator from the one in the numerator.  

f^{9}/f^{3}=f^{9-3}

3 0
3 years ago
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