Which rectangle is similar to a rectangle 4 inches wide and 8 inches long

16 is 4670 of what number ?<br><br><br> Helppp

maxonik [38]

Answer:

16

Step-by-step explanation:

Hope this helps sorry if I'm wrong

8 0

2 years ago

Read 2 more answers

One sample has Aa Aa One sample has n 10 scores and a variance of s2 20, and a second sample has n 15 scores and a variance of s

siniylev [52]

Answer:

option (a) It will be closer to 30 than to 20

Step-by-step explanation:

Data provided in the question:

For sample 1:

n₁ = 10

variance, s₁² = 20

For sample 2:

n₂ = 15

variance, s₂² = 30

Now,

The pooled variance is calculated using the formula,

$S^{2}_{p} = \frac{(n_{1}-1)\times s^{2}_{1} +(n_{2}-1)\times s^{2}_{2}}{n_{1}+n_{2}-2}$

on substituting the given respective values, we get

$S^{2}_{p} = \frac{(10-1)\times 20 +(15-1)\times 30}{10+15-2}$

or

$S^{2}_{p}$ = 26.0869

Hence,

the pooled variance will be closer to 30 than to 20

Therefore,

The correct answer is option (a) It will be closer to 30 than to 20

4 0

2 years ago

What are three expressions equivalent to : 2g+10h?. <br> PLEASE!!!!!

stellarik [79]

Answer:

2(1g + 5h)

(2 x g) + (2 x 5h)

4g + 10h

_______

2

I don't know the options so I just did 3 possible ways it could be equal

8 0

3 years ago

Solve the following equation.

elena-14-01-66 [18.8K]

Answer:

x=6

Step-by-step explanation:

x + 6 = x + x original problem

x + 6 = 2x combine like terms

-x = -x subtract x from both sides to have like terms on each side

6 = x solution

8 0

2 years ago

The owner of an automobile insures it against damage by purchasing an insurance policy with a deductible of 250. In the event th

choli [55]

Answer:

Step-by-step explanation:

From the given information:

The uniform distribution can be represented by:

$f_x(x) = \dfrac{1}{1500} ; o \le x \le \ 1500$

The function of the insurance is:

$I(x) = \left \{ {{0, \ \ \ x \le 250} \atop {x -20 , \ \ \ \ \ 250 \le x \le 1500}} \right.$

Hence, the variance of the insurance can also be an account forum.

$Var [I_{(x}) = E [I^2(x)] - [E(I(x)]^2$

here;

$E[I(x)] = \int f_x(x) I (x) \ sx$

$E[I(x)] = \dfrac{1}{1500} \int ^{1500}_{250{ (x- 250) \ dx$

$= \dfrac{1}{1500 } \dfrac{(x - 250)^2}{2} \Big |^{1500}_{250}$

$\dfrac{5}{12} \times 1250$

Similarly;

$E[I^2(x)] = \int f_x(x) I^2 (x) \ sx$

$E[I(x)] = \dfrac{1}{1500} \int ^{1500}_{250{ (x- 250)^2 \ dx$

$= \dfrac{1}{1500 } \dfrac{(x - 250)^3}{3} \Big |^{1500}_{250}$

$\dfrac{5}{18} \times 1250^2$

∴

$Var {I(x)} = 1250^2 \Big [ \dfrac{5}{18} - \dfrac{25}{144}]$

Finally, the standard deviation of the insurance payment is:

$= \sqrt{Var(I(x))}$

$= 1250 \sqrt{\dfrac{5}{48}}$

≅ 404

4 0

2 years ago