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2 years ago

1/4-5/8n is equivalent to p(2-5n)

Mathematics

2 answers:

jolli1 [7]2 years ago

7 0

Answer:

1/4 - 5/8n = p (2-5n)

a_sh-v [17]2 years ago

5 0

Hey there!

The answer is $\frac{-3}{8}n=2p-5np$

Since they are equivalent to each other, we will set them equal to each other, because equivalent means equal to.

$\frac{1}{4}n-\frac{5}{8}n = p(2-5n)$

Now, we will simplify it. First, we will simplify the left side by subtracting. To do this, we can make $\frac{1}{4}$ into $\frac{2}{8}$ . That gives us:

$\frac{-3}{8}n=p(2-5n)$

Now, we'll simplify the right side by distributing $p$ :

$\frac{-3}{8}n=2p-5np$

Have a super awesome day! :)

You might be interested in

Which expression is equivalent to - 19 - (- 27)

HACTEHA [7]

Answer:

Step-by-step explanation:

-19 - (-27)

-19 + 27

i dont know if this is what you needed, but um yea lol

3 0

3 years ago

Can anyone help? thanks in advance!

Kipish [7]

Answer:

B(2,4)

Step-by-step explanation:

A(-1,-9)

M(0.5,-2.5)

with M we can find x₂ and y₂

x = 0.5

x = x₁+x₂/2

0.5 = (-1 + x₂)/2

(x₂ -1)/2 = 0.5

x₂-1=0.5×2

x₂-1=1

x₂=2

for y₂

y= -2.5

y=(y₁+y₂)/2

y=(-9+y₂)/2

-2.5=(-9+y₂)/2

-9+y₂=-2.5×2

-9+y₂=-5

y₂=-5+9

y₂=4

.: coordinates are B(2,4).

6 0

2 years ago

A simulation was conducted using 10 fair six-sided dice, where the faces were numbered 1 through 6. respectively. All 10 dice we

kompoz [17]

Answer:

C) a sample distribution of a sample mean with n = 10

$\mu_{{\overline}{X}} = 3.5$

and $\sigma_{{\overline}{Y}} = 0.38$

Step-by-step explanation:

Here, the random experiment is rolling 10, 6 faced (with faces numbered from 1 to 6) fair dice and recording the average of the numbers which comes up and the experiment is repeated 20 times.So, here sample size, n = 20 .

Let,

$X_{ij}$ = The number which comes up on the ith die on the jth trial.

∀ i = 1(1)10 and j = 1(1)20

Then,

$E(X_{ij})$ = $\frac {1 + 2 + 3 + 4 + 5 + 6}{6}$

= 3.5 ∀ i = 1(1)10 and j = 1(1)20

and,

$E(X^{2}_{ij}$ = $\frac {1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2}}{6}$

= $\frac {1 + 4 + 9 + 16 + 25 + 36}{6}$

= $\frac {91}{6}$

$\simeq$ 15.166667

so, $Var(X_{ij}$ = $(E(X^{2}_{ij} - {(E(X_{ij})}^{2})$

$\simeq 15.166667 - 3.5^{2}$

= 2.91667

and $\sigma_{X_{ij}}$ = $\sqrt {2.91667}[/tex [tex]\simeq 1.7078261036$

Now we get that,

$Y_{j} = \frac {\sum_{j = 1}^{20}X_{ij}}{20}$

We get that $Y_{j}'s$ are iid RV's ∀ j = 1(1)20

Let, ${\overline}{Y} = \frac {\sum_{j = 1}^{20}Y_{j}}{20}$

So, we get that $E({\overline}{Y}) = E(Y_{j})$

= $E(X_{ij}$ for any i = 1(1)10

= 3.5

and,

$\sigma_{({\overline}{Y})} = \frac {\sigma_{Y_{j}}}{\sqrt {20}} = \frac {\sigma_{X_{ij}}}{\sqrt {20}} = \frac {1.7078261036}{\sqrt {20}} [tex]\simeq 0.38$