Laura spent 4/3 of her savings on a new CD which equivalent fraction shows the amount Laura spent? 1/9, 2/8, 1/3, or 2/3

Find the area of the figure given.

kupik [55]

The answer is C - 63.2 cm

4 0

3 years ago

Read 2 more answers

What is the product of (6x3+3x)(x2+4)

ki77a [65]

If you would like to know what is the product of (6 * x^3 + 3 * x) * (x^2 + 4), you can calculate this using the following steps:

(6 * x^3 + 3 * x) * (x^2 + 4) = 6 * x^3 * x^2 + 6 * 4 * x^3 + 3 * x * x^2 + 3 * 4 * x = 6 * x^5 + 24 * x^3 + 3 * x^3 + 12 * x = 6 * x^5 + 27 * x^3 + 12 * x

The correct result would be 6 * x^5 + 27 * x^3 + 12 * x.

3 0

3 years ago

A ship is approaching a shore and makes an angle of elevation of 11.3o to the top of the light house. Find the distance of the s

Serggg [28]

Answer:

it is very very tall hfhfhfudddhehhd

Step-by-step explanation:

hejrjrneudehshheeundjdjdjsjsjsjdususisisisidusijdjsjejeushd7d jeep has 3AM in its not that serious rn im in the morning princess ❤i love it was the last one of my friends not yours and I love you so much I 3AM you so much and I don't know if you want me tk and I will never be a good friend and he was a prank but I was like that and he was so long for me to be a little sibling and he was so much better than you and he was a little upset caude and he was a little nervous and he just stays awake for me to get my hair done w each other for like that and he was so nervous and he just wanted the same

4 0

3 years ago

The news listed that the al?rage price Of a gallon Of gasoline

Virty [35]

Answer:

b. between $3.04 and $3.36

Step-by-step explanation:

Given,

Price of 1 gallon of gasoline = $3.20

Margin of Error = 5%

We have to find out average cost of gallon of gasoline.

For this, we have to find out 5% of the price of gasoline. Which is calculated by dividing 5 by 100 and then multiply it by 3.20.

Amount of margin of error = $\frac{5}{100}\times3.20 = 0.16$

Hence The margin of error is ±, So we will subtract the amount of margin of error with actual amount of 1 gallon of gasoline. Also we will add the amount of margin of error with actual amount of 1 gallon of gasoline.

Margin of error 1 = $3.20-0.16= \$3.04$

Margin of error 2 = $3.20+0.16= \$3.36$

Hence the Average cost of a gallon of gasoline should be between $3.04 and $3.36.

3 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,

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

Hence, the option which best describes the distribution being simulated is given by,

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

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

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

6 0

2 years ago