Assume that a procedure yields a binomial distribution with a trial repeated n times. Use

A = 2.(1+2+3.....224)+225

Dimas [21]

Step-by-step explanation:

$\bf a=2\cdot (1+2+3....224)+225$

$\bf a=2\cdot \big[(1+224)\cdot 224 : 2\big]+225$

$\bf a=2\cdot \big(225\cdot 224 : 2\big)+225$

$\bf a=2\cdot \big(225\cdot 112\big)+225$

$\bf a=2\cdot 25200+225$

$\bf a=50400+225$

$\bf a=50625$

$\red{\boxed{\bf a=225^{2} \implies ~patrat ~perfect}}$

$\bf \sqrt{a}=\sqrt{225^{2}} \implies \purple{\boxed{\bf a = 225}}$

4 0

3 years ago

You have two cards with a sum of (-12) in your hand.

Vitek1552 [10]

Answer:

you could have a 6 and then a -6

or a 3 and a -3

or 0 and 0

have 2 cards with the same number but one is negative

Step-by-step explanation:

4 0

3 years ago

How to solve this? Please show work. x+3 /4 = x-1 /5

Vinil7 [7]

This will take me a long time, but... $x+ \frac{3}{4} = x- \frac{1}{5} =x+ \frac{3}{4} -x = x- \frac{1}{5} -x = \frac{3}{4} =\frac{1}{5} =\frac{3}{4} - \frac{3}{4} =\frac{1}{5} - \frac{3}{4}
$ $=\frac{-19}{20}$

5 0

3 years ago

Can you use Pythagorean Theorem to find the missing side? Why or why not?

Dvinal [7]

No. You cannot use the Pythagorean theorem to find the missing side, because you can only use The Pythagorean theorem when you are dealing with a right triangle.

4 0

3 years ago

Students in a representative sample of 69 second-year students selected from a large university in England participated in a stu

Serhud [2]

Answer:

95% confidence interval estimate of μ, the mean procrastination scale for second-year students at this terval college is [39.34 , 42.66].

Step-by-step explanation:

We are given that for the 69 second-year students in the study at the university, the sample mean procrastination score was 41.00 and the sample standard deviation was 6.89.

Firstly, the pivotal quantity for 95% confidence interval for the true mean is given by;

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

where, $\bar X$ = sample mean procrastination score = 41

s = sample standard deviation = 6.89

n = sample of students = 69

$\mu$ = population mean estimate

Here for constructing 95% confidence interval we have used One-sample t test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the true mean, $\mu$ is ;

P(-1.9973 < $t_6_8$ < 1.9973) = 0.95 {As the critical value of t at 68 degree

of freedom are -1.9973 & 1.9973 with P = 2.5%}

P(-1.9973 < $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ < 1.9973) = 0.95

P( $-1.9973 \times{\frac{s}{\sqrt{n} } }$ < ${\bar X -\mu}$ < $1.9973 \times{\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.9973 \times{\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.9973 \times{\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ =[ $\bar X-1.9973 \times{\frac{s}{\sqrt{n} } }$ , $\bar X+1.9973 \times{\frac{s}{\sqrt{n} } }$ ]

= [ $41-1.9973 \times{\frac{6.89}{\sqrt{69} } }$ , $41+1.9973 \times{\frac{6.89}{\sqrt{69} } }$ ]

= [39.34 , 42.66]

Therefore, 95% confidence interval estimate of μ, the mean procrastination scale for second-year students at this terval college is [39.34 , 42.66].

5 0

3 years ago