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2 years ago

Para que sirve el álgebra

2 answers:

6 0

El álgebra es un tipo de matemática. Se trata de ecuaciones y resolución de incógnitas. El álgebra se usa para calcular números basados en una ecuación. ¡Espero que esto ayude!

Mumz [18]2 years ago

5 0

Answer:

si mamacita

Step-by-step explanation:

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Solve the simultaneous equations: <br> <img src="https://tex.z-dn.net/?f=y%20%3D%20%20x%5E%7B2%7D%20-%203x%20-%203%20" id="TexFo

Paraphin [41]

I hope this helps you

8 0

3 years ago

What is the median of this data set? {29, 22, 23, 31, 29, 29, 22, 23} Enter your answer in the box.

GaryK [48]

Answer:

The median is 26

Step-by-step explanation:

A "median" is equivalent to, when the terms are arranged in the proper order, whichever term is in the "middle".

So- first, arrange the terms in ascending order.

[22, 22, 23, 23, 29, 29, 29, 31]

There are an even number- eight- terms in total, so we will take the average of the two "middle" terms.

(29+23)/2; 29 and 23 are the "middle" terms, and there are two of them.

(52)/2 = 26

Therefore, the median is 26.

I hope this helped! :)

4 0

2 years ago

Please help!!! <br>solve for x

8_murik_8 [283]

I believe it is -7 but I may be misunderstanding the question

3 0

3 years ago

Read 2 more answers

Solve equation for X<br> c(x-3)-4=b(x+7)

Alenkasestr [34]

Answer:

c(x-3)-4=b(x+7)
(cx-c•3)-4=b(x+7)
cx-3c-4=b(x+7)
cx-3c-4=bx+b•7
-3c-7b-4=bx-cx
x(b-c)=-3c-7b-4
x(b-c)/b-c = -3c-7b-4/b-c

x=-3c+7b+4/b-c

3 0

3 years ago

23. A certain type of gasoline is supposed to have a mean octane rating of at least 90. Suppose measurements are taken on of 5 r

Nostrana [21]

Answer:

a) $p_v =P(Z$

b) Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

c) $P(\bar X >90)=1-P(Z$

$1-P(Z$

d) For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

Step-by-step explanation:

Previous concepts and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

We can calculate the sample mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

$\bar X=88.96$ represent the sample mean

$s=1.011$ represent the sample standard deviation

n=5 represent the sample selected

$\alpha$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 90, the system of hypothesis would be:

Null hypothesis: $\mu \geq 90$

Alternative hypothesis: $\mu < 90$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{88.96-90}{\frac{0.8}{\sqrt{5}}}=-2.907$

Part a

P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=5-1 = 4$

Then since is a left tailed test the p value would be:

$p_v =P(Z$

Part b

Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

Part c

We want this probability:

$P(\bar X$

The best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

If we apply this formula to our probability we got this:

$P(\bar X >90)=1-P(Z$

$1-P(Z$

Part d

For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

7 0

3 years ago

Para que sirve el álgebra​

Para que sirve el álgebra