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2 years ago

6

Which graph be represents the soulution set of y

q" align="absmiddle" class="latex-formula">3/4x-4?

1 answer:

Artemon [7]2 years ago

6 0

The graph would intercept at (0,-4) on the y-axis & at (5.3,0) on the x-axis. And the shaded region would be on the right side of the line on the graph.

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Find the value of k if (x-3) is a factor of P(x)=2x3-5x2+3x+k

blsea [12.9K]

Answer:

k = -18

Step-by-step explanation:

If x-3 is a factor, what this means is that if we substitute;

x = 3 into the equation, we should get 0

This is from the factor theorem

Thus ;

P(3) = 0 = 2(3)^3 -5(3)^2 + 3(3) + k

0 = 54 - 45 + 9 + k

k + 18 = 0

k = -18

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2 years ago

Whats 2.13×0.197 as a decimal

Tcecarenko [31]

Answer:

It's 0.41961.

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2 years ago

Read 2 more answers

Calcula el lado y el área de un poligono regular 12 lados sabiendo que su radio es de 7,727dmy su apotema de 7,464

Vika [28.1K]

Answer:

$L=3.999 \: dm$

$A=183.4581\:dm^{2}$

Step-by-step explanation:

La relación del apotema y el lado de un dodecágono regular está dada por la siguiente ecuación:

$a_{p}=L\frac{2+\sqrt{3}}{2}$

Sabiendo que la apotema es 7.464 el lado sera:

$L=\frac{2*a_{p}}{2+\sqrt{3}}$

$L=\frac{2*7.464}{2+\sqrt{3}}$

$L=3.999 \: dm$

Por otro lado el area de un dodecágono regular esta dada por:

$A=6a_{p}L$

$A=6*7.646*3.999$

$A=183.4581\:dm^{2}$

Espero te haya sido de ayuda!

7 0

2 years ago

For parts a and bâ, use technology to estimate the following. âa) The critical value of t for a 90â% confidence interval with df

rjkz [21]

Answer:

a) For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

$t_{\alpha/2} =\pm 2.35$

b) For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

$t_{\alpha/2} =\pm 2.62$

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

Part a

For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

$t_{\alpha/2} =\pm 2.35$

Part b

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

$t_{\alpha/2} =\pm 2.62$

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2 years ago

HELPPP <br> 8g + 25 = 97<br> Solve for G

RoseWind [281]

Answer: 120 g

Step-by-step explanation:

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1 year ago