HELP ME IT'S URGENT I COULD FAIL

A parallelogram with all angles measuring 90 degrees is what type of quadrilateral

bixtya [17]

I think the correct answer would be a rectangle
I hope this helps !

7 0

4 years ago

Which equation of the least squares regression line most closely matches the data set?

Rudiy27

Answer: y = 3.5 x+ 43.8

Step-by-step explanation:

Here x represents the number of years after 1990

Thus, we get the table that is used to find the equation will be,

x 0 2 4 6 8

y 45 51 57 61 75

Let the equation that shows the above data,

y = b + a x ---------(1)

$\text{Where, }a=\frac{\sum y \sum x^2-\sum x \sum xy}{n(\sum x^2)-(\sum x)^2}$

$\text{And, }b = \frac{\sum xy - \sum x\sum y}{n\sum x^2 - (\sum x)^2}$

By the above table,

$\sum x = 20$

$\sum xy = 1296$

$\sum x^2 = 120$

$\sum y = 289$

By substituting these values in the above value of a and b,

We get b = 43.8 and a = 3.5

Substitute this value in equation (1)

we get, the equation that shows the given data is,

y = 3.5 x + 43.8

⇒ Option (3) is correct.

4 0

4 years ago

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When r = 4, S = 9, and t = 2. * 5r2 + 4s (4 + 6t) - 12 pls help!!

Lina20 [59]

Step-by-step explanation:

5×16+4×9=116

(4+6×2)-12=4

3 0

3 years ago

The age of the children in kindergarten on the first day of school is uniformly distributed between 4.8 and 5.8 years old. A fir

Kazeer [188]

Answer:

(1) (c) 5.30 years.

(2) (b) 0.289.

(3) (b) 0.80.

(4) (d) 0.50.

(5) (a) 5.25 years.

Step-by-step explanation:

Let X = age of the children in kindergarten on the first day of school.

The random variable X follows a continuous Uniform distribution with parameters a = 4.8 years and b = 5.8 years.

The probability density function function of X is:

$f_{X}(x)=\left \{ {{\frac{1}{b-a}} ;\ a$

(1)

The expected value of a Uniform random variable is:

$E(X)=\frac{1}{2}(a+b)$

Compute the mean of X as follows:

$E(X)=\frac{1}{2}(a+b)=\frac{1}{2}\times (4.8+5.8)=5.3$

Thus, the mean of the distribution is (c) 5.30 years.

(2)

The standard deviation of a Uniform random variable is:

$SD(X)=\sqrt{\frac{1}{12}(b-a)^{2}}$

Compute the standard deviation of X as follows:

$SD(X)=\sqrt{\frac{1}{12}(b-a)^{2}}=\sqrt{\frac{1}{12}\times (5.8-4.8)^{2}}=0.289$

Thus, the standard deviation of the distribution is (b) 0.289.

(3)

Compute the probability that a randomly selected child is older than 5 years old as follows:

$P(X>5)=\int\limits^{5.8}_{5} {\frac{1}{5.8-4.8}}\, dx\\$

$=\int\limits^{5.8}_{5} {1}\, dx\\=[x]^{5.8}_{5}\\=(5.8-5)\\=0.8$

Thus, the probability that a randomly selected child is older than 5 years old is (b) 0.80.

(4)

Compute the probability that a randomly selected child is between 5.2 years and 5.7 years old as follows:

$P(5.2$

$=\int\limits^{5.7}_{5.2} {1}\, dx\\=[x]^{5.7}_{5.2}\\=(5.7-5.2)\\=0.5$

Thus, the probability that a randomly selected child is between 5.2 years and 5.7 years old is (d) 0.50.

(5)

It is provided that a randomly selected child is at the 45th percentile.

This implies that:

P (X < x) = 0.45

Compute the value of x as follows:

$P (X < x) = 0.45$

$\int\limits^{x}_{4.8} {\frac{1}{5.8-4.8}}\, dx=0.45$

$\int\limits^{x}_{4.8} {1}\, dx=0.45$

$[x]^{x}_{4.8}=0.45$

$x-4.8=0.45\\$

$x=0.45+4.8\\x=5.25$

Thus, the age of the child at the 45th percentile is (a) 5.25 years.

6 0

3 years ago

I dont get this please help me on this thanks

marishachu [46]

Answer:

x=-2

Step-by-step explanation:

-4(x-7)=36

divide both sides by four

x-7=-9

move constant to the right and change its sign

x=-9+7

calculate the sum

x=-2

7 0

3 years ago

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