Answer correctly please for brainliest

Subtract – 7x + 6 from –2x^2+9x-2

adelina 88 [10]

Answer:

−2x2+16x−8

Step-by-step explanation:

4 0

3 years ago

Un huerto escolar tiene una superficie de 4dam2 500m2 y se requiere dividir en seis parcelas rectangulares de igual area ¿ cuant

vitfil [10]

Answer:

Cada parcela tendrá una extensión de 150 metros cuadrados.

Step-by-step explanation:

De acuerdo con el enunciado el huerto escolar tiene una extensión de 4 decámetros y 500 metros cuadrados.

Tengamos presente que un decámetro es igual a 100 metros cuadrados, entonces el huerto escolar tiene un área total de 900 metros cuadrados. Si dividimos esa área en seis parcelas de igual extensión, tenemos que:

$x = \frac{900\,m^{2}}{6}$

$x = 150\,m^{2}$

Cada parcela tendrá una extensión de 150 metros cuadrados.

7 0

3 years ago

Freeeeeeeeeeee pointttttttttssss!!!!

Aleks [24]

Answer:

Thank you for the free points

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

The quotient of a number X and 2 is 20

pychu [463]

X=10 because 10x2=20

8 0

3 years ago

The tensile strength of a fiber used in manufacturing cloth is of interest to the purchaser. Previous experience indicates that

yaroslaw [1]

Answer:

a) $z=\frac{127-125}{\frac{2}{\sqrt{8}}}=2.828$

Since is a one-side upper test the p value would be:

$p_v =P(z>2.828)=1-P(z$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and we can conclude that the true mean is higher than 125

b) We select a one sided interval since the interest is in analyze if the true mean is higher than 125, and then the alternative hypothesis needs to be the alternative hypothesis.

Step-by-step explanation:

Data given and notation

$\bar X=127$ represent the mean for the sample

$\sigma=2$ represent the standard deviation for the population

$n=8$ sample size

$\mu_o =125$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

Part a

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the true mean is higher than 125:

Null hypothesis: $\mu = 125$

Alternative hypothesis: $\mu > 125$

We know the population deviation, so for this case is better apply a z 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:

$z=\frac{127-125}{\frac{2}{\sqrt{8}}}=2.828$

Calculate the P-value

Since is a one-side upper test the p value would be:

$p_v =P(z>2.828)=1-P(z$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and we can conclude that the true mean is higher than 125

Part b

We select a one sided interval since the interest is in analyze if the true mean is higher than 125, and then the alternative hypothesis needs to be the alternative hypothesis.

5 0

3 years ago