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3 years ago

It is given that D is inversely proportional to v m

Mathematics

1 answer:

Darina [25.2K]3 years ago

8 0

Answer:

Step-by-step explanation:

$D=\frac{k}{\sqrt{m}}\\\\64=\frac{k}{\sqrt{9}}\\\\64=\frac{k}{3}\\\\ 64*3=k\\\\k=192\\a)D=\frac{192}{\sqrt{4}}\\\\=\frac{192}{2}\\\\D=96\\\\b)32=\frac{192}{\sqrt{m}}\\\\ 32*\sqrt{m}=192\\\\\sqrt{m}=\frac{192}{32}=6\\\\m=36$

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Answer:

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Step-by-step explanation:

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3 years ago

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Write 3.005 using words first person is the brainliest

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Answer:

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3 years ago

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1. Solve the equation.

Keith_Richards [23]

Answer:

1. -3

2. -12

3. 20

4. -15

5. 6

6. -15w+5

7. 10

8. C<-26

9.p> -40

10. P> 16

Step-by-step explanation:

~ -9p -17 = 10

-9p = 27

P = -3

~ x/4 -5 = -8

x/4 = -3

x = -12

~ p/5 +6 =10

p/5 = 4

p = 20

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-2m+60=-6m

4m = -60

m = -15

~ 3n +2=8+2n

n = 6

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4 years ago

If a binomial event has a probability of success of 0.4, how many successes would you expect out of 6000 trials?

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3 years ago

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Construct a 99% confidence interval for the population mean, mu. Assume the population has a normal distribution. A group of 1

Zarrin [17]

Answer:

99% confidence interval for the population mean is [19.891 , 24.909].

Step-by-step explanation:

We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.

Assuming the population has a normal distribution.

Firstly, the pivotal quantity for 99% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean age of selected students = 22.4 years

s = sample standard deviation = 3.8 years

n = sample of students = 19

$\mu$ = population mean

Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.878 < $t_1_8$ < 2.878) = 0.99 {As the critical value of t at 18 degree of

freedom are -2.878 & 2.878 with P = 0.5%}

P(-2.878 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 2.878) = 0.99

P( $-2.878 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.878 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ , $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ]

= [ $22.4 -2.878 \times {\frac{3.8}{\sqrt{19} }$ , $22.4 +2.878 \times {\frac{3.8}{\sqrt{19} }$ ]

= [19.891 , 24.909]

Therefore, 99% confidence interval for the population mean is [19.891 , 24.909].

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3 years ago