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timofeeve [1]
3 years ago
9

Help I need help plz

Mathematics
1 answer:
stiv31 [10]3 years ago
4 0

Answer:

So if you know about combining like terms(if not I reccomend watching a video online), then the different variables are x squared, xy, and z. These are all the variables. I hope this helps!

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what are the mean variacne and standard deviatioon of the values? round to the nerates tenth. 11,11,3,1,11
alexdok [17]

Answer:

Mean = 8

Variance = 22

Standard deviation = 4.6904

Step-by-step explanation:

To find out Mean the formula is : total sum of data set / number of values

(11 + 11 + 3 + 1 + 11 + 11) / 6 = 48 / 6 = 8

Mean = 8

Now for variance we will form a table

x                  11     11     3      1      11      11

(x - Mean)    3     3    -5     -7      3      3  

(x - Mean)²   9     9    25    49    9      9

Now we the formula of variance =

Variance = (9+9+25+49+9+9)/(6-1) = 110/5 = 22

Variance = 22

Now we know Standard deviation = √(variance)

Therefore standard deviation = √22 = 4.6904

Hope this helps!<3

4 0
2 years ago
Read 2 more answers
The LCD is a + 1 (a + 1)³ (a + 1)6
Naily [24]

Answer:

1

Step-by-step explanation:

8 0
3 years ago
How many unique planes can be determined by four noncoplanar points?
marta [7]

Answer:

Four unique planes

Step-by-step explanation:

Given that the points are non co-planar, triangular planes can be formed by the joining of three points

The points will therefore appear to be at the corners of a triangular pyramid or tetrahedron such that together the four points will form a three dimensional figure bounded by triangular planes

The number of triangular planes that can therefore be formed is given by the combination of four objects taking three at a time as follows;

₄C₃ = 4!/(3!×(4-3)! = 4

Which gives four possible unique planes.

3 0
3 years ago
A soccer field is a rectangle 60 meters wide and 110 meters long. The coach asks players to run from one corner of the grass fie
Wittaler [7]

Answer:

A

Step-by-step explanation:

A soccer field is a rectangle 60 meters wide and 110 meters long. The coach asks players to run from one corner of the grass field to the opposite corner. The path from one corner of the grass field to the opposite corner is the diagonal of the rectangular field.

The width, the length and the diagonal of the rectangular field form right triangle. By the Pythagorean theorem,

\text{Diagonal}^2=\text{Width}^2+\text{Length}^2\\ \\\text{Diagonal}^2=60^2+110^2\\ \\\text{Diagonal}^2=3,600+12,100\\ \\\text{Diagonal}^2=15,700\\ \\\text{Diagonal}=\sqrt{15,700}\approx 125\text{ yd}

6 0
3 years ago
When fishing off the shores of Florida, a spotted seatrout must be between 10 and 30 inches long before it can be kept; otherwis
Mademuasel [1]

Answer:

There is a 97.585% probability that a fisherman catches a spotted seatrout within the legal limits.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by

Z = \frac{X - \mu}{\sigma}

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

The lengths of the spotted seatrout that are caught, are normally distributed with a mean of 22 inches, and a standard deviation of 4 inches, so \mu = 22, \sigma = 4.

What is the probability that a fisherman catches a spotted seatrout within the legal limits?

They must be between 10 and 30 inches.

So, this is the pvalue of the Z score of X = 30 subtracted by the pvalue of the Z score of X = 10

X = 30

Z = \frac{X - \mu}{\sigma}

Z = \frac{30 - 22}{4}

Z = 2

Z = 2 has a pvalue of 0.9772

X = 10

Z = \frac{X - \mu}{\sigma}

Z = \frac{10 - 22}{4}

Z = -3

Z = -3 has a pvalue of 0.00135.

This means that there is a 0.9772-0.00135 = 0.97585 = 97.585% probability that a fisherman catches a spotted seatrout within the legal limits.

3 0
3 years ago
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