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3 years ago

How do you write 150,000 in scientific notation?

Mathematics

1 answer:

notka56 [123]3 years ago

3 0

Answer:

1.5*10^5

Step-by-step explanation:

So here is what I did, I took the first 2 numbers, then turned it into 1.5,

that leaves us with 4 0's but with the .5 which counts as another one. Now I have 5, which goes as the exponent of 10.

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Addison made 11 quarts of punch for her party. How many gallons of punch did she make? THANK YOU ;)

artcher [175]

11 quarts=2.75 gallons. 4 quarts eqals a gallon. Hope that helps!!!!

8 0

3 years ago

Find the standard equation of the parabola that satisfies the given conditions. Also, find the length of the latus rectum of eac

lakkis [162]

Answer:

The standard parabola

y² = -18 x +27

Length of Latus rectum = 4 a = 18

Step-by-step explanation:

<u><em>Explanation:-</em></u>

Given focus : (-3 ,0) ,directrix : x=6

Let P(x₁ , y₁) be the point on parabola

PM perpendicular to the the directrix L

SP² = PM²

(x₁ +3)²+(y₁-0)² = $(\frac{x_{1}-6 }{\sqrt{1} } )^{2}$

x₁²+6 x₁ +9 + y₁² = x₁²-12 x₁ +36

y₁² = -18 x₁ +36 -9

y₁² = -18 x₁ +27

The standard parabola

y² = -18 x +27

Length of Latus rectum = 4 a = 4 (18/4) = 18

5 0

3 years ago

It takes 18 hours to fill a pool using 8 garden hoses. How long would it take to fill the same pool using 9 garden hoses?

levacccp [35]

Use rates, so 8 garden hoses over 18 hours
then find the how many hours one garden house can fill up by doing this = 18/8 you divide the 8 for both sides which is about 2.25/1
so then you times 9 which is about 20/9 hoses
your answer is 20

3 0

3 years ago

Read 2 more answers

Let X1,X2......X7 denote a random sample from a population having mean μ and variance σ. Consider the following estimators of μ:

Viefleur [7K]

Answer:

a) In order to check if an estimator is unbiased we need to check this condition:

$E(\theta) = \mu$

And we can find the expected value of each estimator like this:

$E(\theta_1 ) = \frac{1}{7} E(X_1 +X_2 +... +X_7) = \frac{1}{7} [E(X_1) +E(X_2) +....+E(X_7)]= \frac{1}{7} 7\mu= \mu$

So then we conclude that $\theta_1$ is unbiased.

For the second estimator we have this:

$E(\theta_2) = \frac{1}{2} [2E(X_1) -E(X_3) +E(X_5)]=\frac{1}{2} [2\mu -\mu +\mu] = \frac{1}{2} [2\mu]= \mu$

And then we conclude that $\theta_2$ is unbiaed too.

b) For this case first we need to find the variance of each estimator:

$Var(\theta_1) = \frac{1}{49} (Var(X_1) +...+Var(X_7))= \frac{1}{49} (7\sigma^2) = \frac{\sigma^2}{7}$

And for the second estimator we have this:

$Var(\theta_2) = \frac{1}{4} (4\sigma^2 -\sigma^2 +\sigma^2)= \frac{1}{4} (4\sigma^2)= \sigma^2$

And the relative efficiency is given by:

$RE= \frac{Var(\theta_1)}{Var(\theta_2)}=\frac{\frac{\sigma^2}{7}}{\sigma^2}= \frac{1}{7}$

Step-by-step explanation:

For this case we assume that we have a random sample given by: $X_1, X_2,....,X_7$ and each $X_i \sim N (\mu, \sigma)$

Part a

In order to check if an estimator is unbiased we need to check this condition:

$E(\theta) = \mu$

And we can find the expected value of each estimator like this:

$E(\theta_1 ) = \frac{1}{7} E(X_1 +X_2 +... +X_7) = \frac{1}{7} [E(X_1) +E(X_2) +....+E(X_7)]= \frac{1}{7} 7\mu= \mu$

So then we conclude that $\theta_1$ is unbiased.

For the second estimator we have this:

$E(\theta_2) = \frac{1}{2} [2E(X_1) -E(X_3) +E(X_5)]=\frac{1}{2} [2\mu -\mu +\mu] = \frac{1}{2} [2\mu]= \mu$

And then we conclude that $\theta_2$ is unbiaed too.

Part b

For this case first we need to find the variance of each estimator:

$Var(\theta_1) = \frac{1}{49} (Var(X_1) +...+Var(X_7))= \frac{1}{49} (7\sigma^2) = \frac{\sigma^2}{7}$

And for the second estimator we have this:

$Var(\theta_2) = \frac{1}{4} (4\sigma^2 -\sigma^2 +\sigma^2)= \frac{1}{4} (4\sigma^2)= \sigma^2$

And the relative efficiency is given by:

$RE= \frac{Var(\theta_1)}{Var(\theta_2)}=\frac{\frac{\sigma^2}{7}}{\sigma^2}= \frac{1}{7}$

5 0

4 years ago

Please help I will give Brainliest please. Just tell what I need to do or solve I need help ASAP *image*

Ksju [112]

Answer:

Step-by-step explanation:

Hard to figure out huh

8 0

3 years ago