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2 years ago

Answer this please please

Mathematics

1 answer:

zloy xaker [14]2 years ago

6 0

Answer:

x is less than -3

Step-by-step explanation:

3x^2+9x<0

x^2+3x<0

x(x+3)<0

x+3<0

x<-3

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Dante runs 7 miles in 60 minutes. At the same rate, how many miles would he run in 24 minutes?

denis23 [38]

<h2>Answer: 2.8 miles</h2>

Step-by-step:

We can write it in the proportion $\frac{7}{60} = \frac{x}{24}$

where x is our number of miles trying to find

We now cross multiply or use the butterfly method which I will demonstrate both

<h3>Cross Multiply</h3>

We multiply the numerator with the dominator

so $7 multiplied by 24$

and $60 multiplied by x$

$7*24=60x$

$168=60x$

$\frac{168}{60} =\frac{60x}{60}$

$2.8=x$ or $x=2.8$

Now for the

<h3>Butterfly Method </h3>

You would do 7 times 24 divided by 60 or $\frac{7*24}{60}$

Then $\frac{168}{60}$ simplify

$2.8$ so 2.8 miles in either method

3 0

3 years ago

Which value of K will make x^2-1/4x+k a perfect square trinomial?

lakkis [162]

To find it it is
ax^2+bx+c
if the first term (a) is 1 then c=b/2
b=-1/4
(-1/4)/2=-1/4 times 1/2=-1/8

k=-1/8

4 0

4 years ago

3. The masses of two gorillas are given below. A female gorila has a mass of

loris [4]

Answer: 135,000g

Hope this helped :)

6 0

3 years ago

∠45° is complementary to which of the following angles?

mixas84 [53]

Answer:

45°

Step-by-step explanation:

complementary angles add up to 90° so it's 90-45

4 0

2 years ago

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