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Fofino [41]
4 years ago
12

D=(ry + r) t Solve for r1

Mathematics
2 answers:
TiliK225 [7]4 years ago
3 0

Answer:

i

Step-by-step explanation:

vodka [1.7K]4 years ago
3 0

Answer: r1 = d/t - r2

Step-by-step explanation:

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Evaluate v 49t where t = 4.
const2013 [10]

Answer:

196

Step-by-step explanation:

49(4) = 196

3 0
4 years ago
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I need help guys this time it is 6 points cuz I'm poor so PLZ help I really need it !!!!!!!!
riadik2000 [5.3K]

Answer:

Answer is option d mark as brainiest give 100 point

8 0
3 years ago
The amount of money spent on textbooks per year for students is approximately normal.
Ostrovityanka [42]

Answer:a

a

   336.04    <  \mu < 443.96

b

  The  margin of error will increase

c

The  margin of error will decreases

d

The 99% confidence interval is  0.4107 <  p  < 0.4293

Step-by-step explanation:

From the question we are  told that

   The sample size  n =  19

    The sample mean is  \= x  = \$\  390

    The  standard deviation is  \sigma =  \$ \  120

 

Given that the confidence level is  95% then the level of significance is mathematically represented as

           \alpha = 100 -  95

          \alpha  =  5 \%

          \alpha  =  0.05

Next we obtain the critical value of \frac{\alpha }{2} from the normal distribution table

    So  

         Z_{\frac{\alpha }{2} } =  1.96

The  margin of error is mathematically represented as

      E =  Z_{\frac{\alpha }{2} } *  \frac{\sigma}{\sqrt{n} }

=>    E = 1.96 *  \frac{120}{\sqrt{19} }

=>   E = 53.96

The 95% confidence interval is  

     \= x  -  E  <  \mu < \= x  +  E

=>   390  -   53.96   <  \mu < 390  -   53.96

=>  336.04    <  \mu < 443.96

When the confidence level increases the Z_{\frac{\alpha }{2} } also increases which increases the margin of error hence the confidence level becomes wider

Generally the sample size mathematically varies with margin of error as follows

         n  \  \ \alpha  \ \  \frac{1}{E^2 }

So if the sample size increases the margin of error decrease

The  sample proportion is mathematically represented as

       \r p  =  \frac{210}{500}

       \r p  = 0.42

Given that the confidence level is 0.99 the level of significance is  \alpha =  0.01

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

      Z_{\frac{\alpha }{2} }  =  2.58

  Generally the margin of error is mathematically represented as

       E =  Z_{\frac{\alpha }{2} }*  \sqrt{ \frac{\r p (1- \r p )}{n} }

=>   E =  0.42 *  \sqrt{ \frac{0.42 (1- 0.42 )}{ 500} }

=>     E =  0.0093

The 99% confidence interval  is

     \r p  -  E <  p  < \r p  +  E

     0.42  -  0.0093 <  p  < 0.42  +  0.0093

     0.4107 <  p  < 0.4293

 

4 0
4 years ago
Help help (12 points)
iren2701 [21]

Answer:

2m+2m

Step-by-step explanation:

7 0
3 years ago
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three numbers are in an arithmetic progression. Their sum is 3 and the sum of their squares is 11. what are the three numbers
Vlad [161]

The three numbers are 12, 18 and 24

Arithmetic progression

Let the 3 number in arithmetic progression be:

a-d, d, a+d ...

If their sum is 3, then;

a-d+d+a+d = 3

2a + d = 3 ........... 1

If the sum of their squares is 11, then;

(a-d)² + d² + (a+d)² = 11

a²-2ad+d²+d²+a²+2ad+d²  11

2a²+3d² = 11 ....... 2

Solving the equations simultaneously, d = 6 and a = 12

First-term = 12

second term = 18

Thirs term = 24

Hence the three numbers are 12, 18 and 24

Hope this helps you!!!!!! :D

5 0
3 years ago
Read 2 more answers
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