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2 years ago

Please help i need to bring my grade up :(

Mathematics

1 answer:

VLD [36.1K]2 years ago

4 0

The answer is true
.12 can be written as 12/100 and can be reduced 3/25

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3 3/4 + 2 1/8 = (make sure you simplify the fraction)

dusya [7]

This is 5 7/8 or 47/8

5 0

2 years ago

Read 2 more answers

Can someone help me with number 10, 11, 12, 15, and 16 I don’t understand them thanks

yanalaym [24]

10)

$10^2-48:6+25*3=100-8+75=167$

11)

$8(\frac{16}{4} )+2^2-11*3=32+4-33=3$

12)

$(\frac{6}{3} +4)^2:4*7=(2+4)^2:4*7=6^2:4*7=36:4*7=9*7=63$

13) (your answer is incorrect)

$5(9-4)^2-3^2=5*5^2-3^2=5*25-9=125-9=116$

14) (your answer is incorrect)

$5^2-2^2*4^2-12=25-4*16-12=25-64-12=-51$

15)

$(\frac{50}{5^2} )^2:4=(\frac{50}{25} )^2:4=2^2:4=4:4=1$

16)

$\frac{24+32+30+28}{2}$ -expression to represent this situation

$\frac{24+32+30+28}{2}=\frac{114}{2}=57$

Answer: In 1 group 57 students

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P.S. Hello from Russia :^)

4 0

2 years ago

A researcher plans to conduct a significance test at the 0.01 significance level. She designs her study to have a power of 0.90

Rasek [7]

Answer: 0.10

Step-by-step explanation: The type 2 error is committed when the alternative hypothesis is rejected when it should have been accepted causing the researcher to accept the null hypothesis which is false.

Power is the probability of avoiding a type 2 error. That is ;

Power = 1 - P(type 2 error)

Given that power = 0.90 ; P(type 2 error) = probability of committing a type 2 error.

P(type 2 error)' = 1 - P(type 2 error) = Probability of not committing or avoiding a type 2 error

0.90 = 1 - P(type 2 error)

P(type 2 error) = 1 - 0.90

P(type 2 error) = 0.10

6 0

3 years ago

Pls help if pls do most important 15 number 20 number 26 number 28 number 30 number 17 number 25 number 22 number 24 number if u

Tamiku [17]

Answer:

Step-by-step explanation:

15. $\frac{cotx+1}{cotx-1} = \frac{cotx+cotx.tanx}{cotx-cotx.tanx} = \frac{cotx(1+tanx)}{cotx(1-tanx)} = \frac{1+tanx}{1-tanx} \\$

16. $\frac{1+cos\alpha }{sin\alpha } = \frac{sin\alpha }{1-cos\alpha }$

$=> (1+cos\alpha )(1-cos\alpha ) = sin^{2} \alpha \\=> 1-cos^{2}\alpha =sin^{2} \alpha \\=> sin^{2}\alpha +cos^{2}\alpha =1\\$

=> The clause is correct

17. Do the same (16)

18. $\frac{sinx}{1-cosx}+\frac{sinx}{1+cosx} = \frac{sinx(1+cosx) + sinx(1-cosx)}{(1+cosx)(1-cosx)} = \frac{sinx + sinx.cosx + sinx - sinx.cosx}{1-cos^{2}x} = \frac{2sinx}{sin^{2}x }= \frac{2}{sinx} = 2cosecx$

19.

$\frac{sinA}{1+cosA} + \frac{1+cosA}{sinA}=\frac{2}{sinA} \\=> \frac{sinA}{1+cosA} + \frac{1+cosA}{sinA} - \frac{2}{sinA} \\ = 0\\=> \frac{sinA}{1+cosA} + (\frac{1+cosA}{sinA} - \frac{2}{sinA} \\) = 0\\=> \frac{sinA}{1+cosA} + \frac{1+cosA-2}{sinA} = 0\\=> \frac{sinA}{1+cosA} +\frac{cosA-1}{sinA} = 0\\=> \frac{sin^{2} A+(cosA-1)(1+cosA)}{(1+cosA)sinA}=0\\=> \frac{sin^{2} A+cos^{2} A-1}{(1+cosA)sinA}=0\\=> \frac{0}{(1+cosA)sinA} =0\\$