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murzikaleks [220]
3 years ago
15

I promise brainliest and a exter 25 poinst to the first to answer What is the solution to the inequality 2x ≥ -4? Click the numb

er line until the correct answer is shown...

Mathematics
1 answer:
AlexFokin [52]3 years ago
4 0

:

Answer:

the answer is the arrow going to the right because its not a negative number and a closed circle

Step-by-step explanation:

so that means that 2x is at LEASt more than -4 opposed to this sign> wich just means greater than

because when you work with variables you usually cannot find the exact amount especially when you are rounding so you know that its biigger than no less than or at least -4

the answer is the arrow going to the right because its not a negative number and a closed circle

please please mark as brainliest

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Find the distance around (perimeter) the figure on the left.<br> The perimeter is ___ in.
Phoenix [80]
4 7/8 = 4 14/16 = 78/16
2 7/16 = 39/16
39/16 + 39/16 + 39/16 + 39/16 = 156/16
78/16 + 78/16 = 156/16
156/16 + 156/16 = 312/16 = 19 8/16 = 19 1/2

Answer: 19 1/2 inches


(hope that helped <3)
3 0
2 years ago
Lauren wants to save 30% of her weekly paycheck. How much will she save each week if her paycheck is $150 a week?
Vlad1618 [11]
She will save $45 dollars a week the working is
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3 0
3 years ago
If Aditya can travel 80 km in 2 hours how many kilometres can he travel in 3 and half hours​
ivann1987 [24]

Answer:

distance covered in 2 hours = 80 km

distance covered in 1 hour = 80÷2 = 40 km

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= 140 km

(3 1/2hours = 7/2 hours)

8 0
2 years ago
Factor completely<br> 9-6a-24a^2
Natasha_Volkova [10]
-24a²- 6a +9 = - 3(8a² +6a - 9)

8a² +6a - 9 =0

x=(-b +/-√(b² - 4ac))/2a
x = (-6 +/-√(36+4*8*9)  /(2*8) = (-6 +/-√(324)  /16 =  (-6 +/-18)/16
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-24a²- 6a +9 = - 3(8a² +6a - 9) = -3(a - 0.75)(a + 1.5)
7 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
3 years ago
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