All categories

2 years ago

Write an equation of the line that passes

Mathematics

1 answer:

BaLLatris [955]2 years ago

7 0

Answer: y = 3x - 20

Step-by-step explanation: Start with the equation y=mx+b. Remember that the slope of the parallel line, m, will be the same. Now you have y = 3x + b. Next you have to find b, the y intercept. You can do so by using the point given, (6,-2).

-2 = (3) * (6) + b

b = -20

y=mx+b y = 3x - 20

You might be interested in

The half-life of a radioactive isotope is the time it takes for a quantity of the isotope to be reduced to half its initial mass

Sauron [17]

Answer:

Quantity remaining= 27.5 grams

Step-by-step explanation:

If the initial quantity of the radioactive isotope= 220 grams

And for each half life it will reduce to half of its original quantity

For first half life

Quantity= 220/2

Quantity= 110 grams

Now for the second half life , the quantity to work with will be 110 grams

Second half life

Quantity= 110/2

Quantity= 55 grams

For the third half life , the quantity to work with is 55 grams

Third half life

Quantity= 55/2

Quantity= 27.5 grams

So after 3 half life's of this radioactive isotope, the amount remaining out of 220 = 27.5 grams.

Another simple way to do this

Since half life reduce the isotope by half of its original.

The third half life will reduce it by (1/2)^3= 1/8

So the quantity remaining= 1/8*220

Quantity remaining= 27.5 grams

6 0

2 years ago

A web site was hit 300 times over a period of 15 days Show that over some period of 3 consecutive days, it was hit at least 60 t

marshall27 [118]

Answer:

We will divide the 15 days in five periods of 3 consecutive days each.

Now to solve this we will use the pigeonhole principle.

This states that if (N+1) pigeons occupy N holes, then some hole must have at least 2 pigeons.

So, we have n=300 pigeons and k=5 holes.

$[\frac{n}{k} ]=[\frac{300}{5} ]$

Hence, there is a period of 3 consecutive days in which the website was hit at least 60 times.

6 0

3 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,

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