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3 years ago

Often, anomalous objects are known as _____, since on a scatter plot of the data, they lie far away from other data points.

Mathematics

1 answer:

ycow [4]3 years ago

4 0

Answer:

outliers

Step-by-step explanation:

An outlier is a data point that is significantly different from other observations. An outlier might be due to inconsistency in measurements, or due to an error introduced into the experiment. Outliers cans lie extremely high or low of other observation in statistics, and they usually create a big problem for proper analysis.

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What is the coefficient of the x-term in the expression 4x+5xy-2y

solmaris [256]

Answer:

Step-by-step explanation:

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3 years ago

Helpppp? PLease im very confused

kompoz [17]

Answer:

Yes

Step-by-step explanation:

There is a linear relationship, so there is a constant change in the graph and thus a proportional relationship.

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4 years ago

IF S.P. = Rs 1,869, loss percent = 11% find C.P.

Veseljchak [2.6K]

Here,

Selling Price (S.P) = Rs 1869

Loss = 11%

Cost Price (C.P) = ?

Let, C.P be x

Now,

Sp = Cp - loss% of Cp

1869 = x - 11/100*x

1869 = (100x - 11x)/100

186900 = 89x

◆ x = Rs 2100

Cp = Rs 2100

I hope you understand...

It's a Right answer...

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7 0

3 years ago

The variance in a production process is an important measure of the quality of the process. A large variance often signals an op

Flauer [41]

Answer:

There is a sufficient evidence to support the that machine 1 has the greater variance.

Step-by-step explanation:

The question with the complete details can be found online.

Start by stating the hypotheses.

The null hypothesis states that the variance of machine 1 is less than or equal to machine 2

So, we have:

$H_o: \sigma_1 ^2 \le \sigma_2 ^2$

The alternate hypothesis will then be:

$H_a: \sigma_1 ^2 > \sigma_2 ^2$

So, we have:

$n_1 = 25$ $n_1 = 22$

Calculate the mean of Machine 1 and 2

The mean is:

$\bar x =\frac{\sum x}{n}$

For machine 1, we have:

$\bar x_1 =\frac{2.95+3.45+3.5+.....+3.12}{25}$

$\bar x_1 =\frac{83.21}{25}$

$\bar x_1 =3.3284$

For machine 2, we have:

$\bar x_2 = \frac{3.22 + 3.3 + 3.34 + ..... +3.33}{22}$

$\bar x_2 = \frac{72.12}{22}$

$\bar x_2 = 3.2782$

Calculate the standard deviation of both machines

The standard deviation is:

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}$

For machine 1, we have:

$\sigma_1 = \sqrt{\frac{(2.95 - 3.3284)^2+(3.45 - 3.3284)^2+(3.5 - 3.3284)^2+.....+(3.12 - 3.3284)^2}{25-1}}$

$\sigma_1 = \sqrt{\frac{(2.95 - 3.3284)^2+(3.45 - 3.3284)^2+(3.5 - 3.3284)^2+.....+(3.12 - 3.3284)^2}{24}}$

$\sigma_1 = 0.2211$

For machine 2, we have:

$\sigma_2 = \sqrt{\frac{(3.22 - 3.2782)^2+(3.3 - 3.2782)^2+(3.44 - 3.2782)^2+.....+(3.33 - 3.2782)^2}{22-1}}$

$\sigma_2 = \sqrt{\frac{(3.22 - 3.2782)^2+(3.3 - 3.2782)^2+(3.44 - 3.2782)^2+.....+(3.33 - 3.2782)^2}{21}}$

$\sigma_2 = 0.0768$

Calculate the degrees of freedom

$df=n-1$

For machine 1;

$df_1=25-1=24$

For machine 2

$df_2=22-1=21$

Calculate the test statistic (t)

$t = \frac{Var_1}{Var_2}$

Rewrite in terms of standard deviation

$t = \frac{\sigma_1^2}{\sigma_2^2}$

$t = \frac{0.2211^2}{0.0768^2}$

$t = \frac{0.04888521}{0.00589824}$

$t = 8.2881$

Lastly, calculate the p value.

This is the value of P(t > 8.2881) between the degrees of freedom i.e. 21 and 24

From the f distribution table

$P(t > 8.2881) < 0.01$

Hence, we reject the null hypothesis

3 0

3 years ago

Variation help please (last one) picture down below:)

Alex Ar [27]

IT IS A 49.6/4=12.4 ZEBRA

3 0

3 years ago