Hi, prove that (AXB).(CXD)+(BXC).(AXD)+(CXA).(BXD)=0

Lance has three times as many pencils as Nick, and they have 84 pencils together. How many pencils does each of them have?

hoa [83]

Is 47 l think !!!!!!!!!!!!!!!!!!!!!

6 0

2 years ago

Rewrite (2x + 8) + 4 using the associative property. Then simplify.

skelet666 [1.2K]

Answer:

2x + 12

Step-by-step explanation:

Example of Associative Property Addition:

a + (b+c) = b + (a+c)

Now, implement it to the given expression:

(2x+8) + 4 = 2x + (8+4)

That simplifies to:

2x + 12

4 0

2 years ago

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The student government association at a certain college assigns a number to each student enrolled, puts the numbers in order, an

snow_tiger [21]

Answer: Systematic

Step-by-step explanation:

3 0

2 years ago

Which are the solutions of x2 = –5x + 8? StartFraction negative 5 minus StartRoot 57 EndRoot Over 2 EndFraction comma StartFract

serg [7]

Answer:

$x=\frac{-5-\sqrt{57} }{2}\ or\ x=\frac{-5+\sqrt{57} }{2}$

Step-by-step explanation:

Given:

The equation to solve is given as:

$x^2=-5x+8$

Rearrange the given equation in standard form $ax^2+bx +c =0$ , where, $a,\ b,\ and\ c$ are constants.

Therefore, we add $5x-8$ on both sides to get,

$x^2+5x-8=0$

Here, $a=1,b=5,c=-8$

The solution of the above equation is determined using the quadratic formula which is given as:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Plug in $a=1,b=5,c=-8$ and solve for $x$ .

$x=\frac{-5\pm \sqrt{5^2-4(1)(-8)}}{2(1)}\\x=\frac{-5\pm \sqrt{25+32}}{2}\\x=\frac{-5\pm \sqrt{57}}{2}\\\\\\\therefore x=\frac{-5-\sqrt{57} }{2}\ or\ x=\frac{-5+\sqrt{57} }{2}$

Therefore, the solutions are:

$x=\frac{-5-\sqrt{57} }{2}\ or\ x=\frac{-5+\sqrt{57} }{2}$

4 0

2 years ago

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Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by $2P(T$ 0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}$ , i.e.,

$-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}$

where $t_{0.025}$ is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

$-0.03\pm(2.0167)(0.012459)(0.3)$ , i.e.,

(-0.0225, -0.0375)

8 0

2 years ago