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Helga [31]
3 years ago
7

The square of a number x is 115. The number x is between what two consecutive whole numbers?

Mathematics
1 answer:
Kipish [7]3 years ago
4 0

Answer:

10 and 11

Step-by-step explanation:

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Advocard [28]
The area of the region on the map is 97.5 miles
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2 years ago
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write the equation of a line that includes the point (1,5) and has a slope of 3 in slope intercept form.
lina2011 [118]
Y=3x+? (1,5)
5=3(1)+?
5=3+?

y=3x+2

7 0
3 years ago
Please Help <br> Write a system of linear Inequalities to represent each graph.
docker41 [41]

Answer:

\left\{\begin{array}{l}5x-y-7\le 0\\2x+3y-9\ge 0\end{array}\right.

Step-by-step explanation:

1. Blue region. The boundary blue line passes through the points (0,-7) and (3,8), then its equation is

\dfrac{x-0}{3-0}=\dfrac{y+7}{8+7},\\ \\15x=3(y+7),\\ \\5x=y+7,\\ \\5x-y-7=0.

This line is solid, so the sign for the inequality should be with the notion "or equal to".

From the diagram you can see that the origin belongs to the blue region, then its coordinates satisfy the inequality. Thus,

5x-y-7\le 0.

2. Green region. The boundary green line passes through the points (6,-1) and (0,3), then its equation is

\dfrac{x-0}{6-0}=\dfrac{y-3}{-1-3},\\ \\-4x=6(y-3),\\ \\-2x=3y-9,\\ \\2x+3y-9=0.

This line is solid, so the sign for the inequality also should be with the notion "or equal to".

From the diagram you can see that the origin doesn't belong to the blue region, then its coordinates satisfy the inequality. Thus,

2x+3y-9\ge 0.

3. The system of linear inequalities that represents these graphs is

\left\{\begin{array}{l}5x-y-7\le 0\\2x+3y-9\ge 0\end{array}\right.

6 0
3 years ago
The amount Troy charges to mow a lawn is proportional to the time it takes him to mow the lawn. Troy charges $30 to mow a lawn t
Inessa05 [86]

Answer:

D = 20H

Step-by-step explanation:

The amount Troy (D, in dollars) charges to mow a lawn is proportional to the time (H, in hours) it takes him to mow the lawn.

Hence, we can write D = kH ......... (1), where k is a constant.

Now, given that Troy charges $30 to mow a lawn that took him 1.5 hours to mow.

So, from equation (1) we get  

30 = 1.5k

⇒ k = 20.

Therefore, equation (1) becomes D = 20H (Answer)

3 0
3 years ago
Let X1 and X2 be independent random variables with mean μand variance σ².
My name is Ann [436]

Answer:

a) E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu

So then we conclude that \hat \theta_1 is an unbiased estimator of \mu

E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu

So then we conclude that \hat \theta_2 is an unbiased estimator of \mu

b) Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}

Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}

Step-by-step explanation:

For this case we know that we have two random variables:

X_1 , X_2 both with mean \mu = \mu and variance \sigma^2

And we define the following estimators:

\hat \theta_1 = \frac{X_1 + X_2}{2}

\hat \theta_2 = \frac{X_1 + 3X_2}{4}

Part a

In order to see if both estimators are unbiased we need to proof if the expected value of the estimators are equal to the real value of the parameter:

E(\hat \theta_i) = \mu , i = 1,2

So let's find the expected values for each estimator:

E(\hat \theta_1) = E(\frac{X_1 +X_2}{2})

Using properties of expected value we have this:

E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu

So then we conclude that \hat \theta_1 is an unbiased estimator of \mu

For the second estimator we have:

E(\hat \theta_2) = E(\frac{X_1 + 3X_2}{4})

Using properties of expected value we have this:

E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu

So then we conclude that \hat \theta_2 is an unbiased estimator of \mu

Part b

For the variance we need to remember this property: If a is a constant and X a random variable then:

Var(aX) = a^2 Var(X)

For the first estimator we have:

Var(\hat \theta_1) = Var(\frac{X_1 +X_2}{2})

Var(\hat \theta_1) =\frac{1}{4} Var(X_1 +X_2)=\frac{1}{4} [Var(X_1) + Var(X_2) + 2 Cov (X_1 , X_2)]

Since both random variables are independent we know that Cov(X_1, X_2 ) = 0 so then we have:

Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}

For the second estimator we have:

Var(\hat \theta_2) = Var(\frac{X_1 +3X_2}{4})

Var(\hat \theta_2) =\frac{1}{16} Var(X_1 +3X_2)=\frac{1}{4} [Var(X_1) + Var(3X_2) + 2 Cov (X_1 , 3X_2)]

Since both random variables are independent we know that Cov(X_1, X_2 ) = 0 so then we have:

Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}

7 0
3 years ago
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