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castortr0y [4]
3 years ago
12

Find ab if the coordinates of a are (3, 5 ( and b are (-3, -3)a- 8b- 6c- 12d- 10

Mathematics
1 answer:
strojnjashka [21]3 years ago
6 0
If you use the distance formula you get 10.
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4d. Health Survey. In a survey of 1020 adults in the United States, 44% said that they wash their hands after riding public tran
AleksandrR [38]

Answer:

Discrete. See explanation below

Step-by-step explanation:

We need to remember some previous concepts:

We have two types of numerical data: Discrete and Continuous

When we say Discrete data we are refering to data that is countable or can be expressed with integers in a domain.

In the other case when we talk about continuous data we are refering to data that is continuous in a specified domain, it can contain decimals or rational numbers in the Real numbers for example.

For this special case we know that they select a sample size of n=1020 and the sample proportion of people in the United States who wash their hands after riding public transportation was 0.44 or 44% in percentage.

\hat p = \frac{X}{1020}=0.44

X=0.44*1020=448.8 \approx 449

But the number of subjects on this survey needs to be Discrete, since the possible values are 0,1,2,3,4,.....,n and never we have decimals or continuous data in order to express this.  

3 0
3 years ago
If you see these clouds in the sky, which type of weather might you expect?
UNO [17]
The answer to your question is thunderstorms.
The picture is filled with a lot of dark clouds, in which there will be a probability for thunderstorms to occur. We all know that when there is lightning, thunder will occur after it. With the photo given, there will be the probability of thunderstorm with heavy rain.
6 0
3 years ago
PLEASE HELP
Andrej [43]

Answer: -3 and 0

Step-by-step explanation: f(x) = g(x) where they intersect. Two lines intersect when they have the same x and y values.

5 0
3 years ago
PLEASE HELP ME I WILL GIVE YOU BRAINLIST AND ALL MY POINTS
Alex_Xolod [135]

Answer:

b = y-intercept; The equation is y = mx + b. The x and y variables remain as letters, but m and b are replaced by numbers (ex: y = 2x + 4, slope = 2 and y-intercept = 4). The following video will show a few examples of understanding how to use the slope and intercept from an equation.

Vertex (4, -13) y = x^2 - 8x + 3 x-coordinate of vertex: x = -b/(2a) = 8/2 = 4 y-coordinate of vertex: y(4) = 16 - 32 + 3 = -13 Vertex (4, -13) To find y-intercepts, make x = 0 --> y = 3 To find x-intercepts, solve the quadratic equation y = 0 Use the improved quadratic formula D = d^2 = b^2 - 4ac = 64 - 12 = 52 --> d = +- 2sqrt13 There are 2 x-intercepts (2 real roots): x = -b/(2a) +- d/(2a) = 8/2 +- (2sqrt13)/2 = 4 +- sqrt13 graph{x^2 - 8x + 3 [-40, 40, -20, 20]}

Step-by-step explanation:

7 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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