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3 years ago

7

Let p be the plane in space that intersects the x-axis at −5, the y-axis at −2, and the z-axis at 5. Find a vector v¯¯¯ that is

perpendicular to p.

1 answer:

Lana71 [14]3 years ago

7 0

Let P be the plane that intersects

x-axis at point (-5,0,0);
y-axis at point (0,-2,0);
z-axis at point (0,0,5).

Write the equation of the plane P:

$\left|\begin{array}{ccc}x-(-5)&y-0&z-0\\0-(-5)&-2-0&0-0\\0-(-5)&0-0&5-0\end{array}\right|=0\Rightarrow \left|\begin{array}{ccc}x+5&y&z\\5&-2&0\\5&0&5\end{array}\right|=0.$

Then

$-10(x+5)+10z-25y=0,\\ \\2(x+5)-2z+5y=0,\\ \\2x+5y-2z=-10.$

The coefficients at variables x, y and z are the coordinates of perpendicular vector to the plane. Thus

$\vec{v}=(2,5,-2)\perp P.$

Answer: $\vec{v}=(2,5,-2).$

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A candy bar box is in the shape of a triangular prism. The volume of the box is 1,200 cubic centimeters.

marin [14]

Helloooo, Your answer for part A is 13 cm

You answer for part b is 2600

13 x 10 x 20 = 2600

3 0

2 years ago

Triangle ABC has vertices of A(-6,7) B(4,-1), and C(-2,-9). Find the length of the median from angle B in triangle ABC.

AfilCa [17]

The length of the median from angle B is 8.

4 0

3 years ago

Plz help me with this

Mkey [24]

Answer: B) A = 750(1.04)ⁿ

<u>Step-by-step explanation:</u>

The formula for compounded annually is: A = P(1 + r)ⁿ where

A (amount accrued) = <em>unknown</em>
P (amount invested) = $750
r (interest rate) = 4% -->(0.04)
t (time in years) = <em>unknown</em>

A = 750(1 + 0.04)ⁿ

= 750(1.04)ⁿ

8 0

2 years ago

A restaurant chain has two locations in a medium-sized town and, believing that it has oversaturated the market for its food, is

scoray [572]

Answer:

Yes, since the test statistic value is greater than the critical value.

Step-by-step explanation:

Data given and notation

$\bar X_{1}=360000$ represent the mean for the downtown restaurant

$\bar X_{2}=340000$ represent the mean for the freeway restaurant

$s_{1}=50000$ represent the sample standard deviation for downtown

$s_{2}=40000$ represent the sample standard deviation for the freeway

$n_{1}=36$ sample size for the downtown restaurant

$n_{2}=36$ sample size for the freeway restaurant

t would represent the statistic (variable of interest)

$\alpha=0.05$ significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the true mean of revenue for downtown is higher than for freeway restaurant, the system of hypothesis would be:

Null hypothesis: $\mu_{1}\leq \mu_{2}$

Alternative hypothesis: $\mu_{1} > \mu_{2}$

Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Determine the critical value.

Based on the significance level $\alpha=0.05$ and $\alpha/2=0.025$ we can find the critical values from the t distribution dith degrees of freedom df=36+36-2=55+88-2=70, we are looking for values that accumulates 0.025 of the area on the right tail on the t distribution.

For this case the value is $t_{1-\alpha/2}=1.66$

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

$t=\frac{360000-340000}{\sqrt{\frac{50000^2}{36}+\frac{40000^2}{36}}}}=1.874$

What is the p-value for this hypothesis test?

Since is a right tailed test the p value would be:

$p_v =P(t_{70}>1.874)=0.033$

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the true mean for the downtown revenue restaurant seems higher than the true mean revenue for the freeway restaurant.

The best option would be:

Yes, since the test statistic value is greater than the critical value.

5 0

2 years ago

In a certain assembly plant, three machines B1, B2, and B3, make 30%, 20%, and 50%, respectively. It is known from past experien

diamong [38]

Answer:

The probability that a randomly selected non-defective product is produced by machine B1 is 11.38%.

Step-by-step explanation:

Using Bayes' Theorem

P(A|B) = $\frac{P(B|A)P(A)}{P(B)} = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|a)P(a)}$

where

P(B|A) is probability of event B given event A

P(B|a) is probability of event B not given event A

and P(A), P(B), and P(a) are the probabilities of events A,B, and event A not happening respectively.

For this problem,

Let P(B1) = Probability of machine B1 = 0.3

P(B2) = Probability of machine B2 = 0.2

P(B3) = Probability of machine B3 = 0.5

Let P(D) = Probability of a defective product

P(N) = Probability of a Non-defective product

P(D|B1) be probability of a defective product produced by machine 1 = 0.3 x 0.01 = 0.003

P(D|B2) be probability of a defective product produced by machine 2 = 0.2 x 0.03 = 0.006

P(D|B3) be probability of a defective product produced by machine 3 = 0.5 x 0.02 = 0.010

Likewise,

P(N|B1) be probability of a non-defective product produced by machine 1 = 1 - P(D|B1) = 1 - 0.003 = 0.997

P(N|B2) be probability of a non-defective product produced by machine 2 = 1 - P(D|B2) = 1 - 0.006 = 0.994

P(N|B3) be probability of a non-defective product produced by machine 3 = 1 - P(D|B3) = 1 - 0.010 = 0.990

For the probability of a finished product produced by machine B1 given it's non-defective; represented by P(B1|N)

P(B1|N) = $\frac{P(N|B1)P(B1)}{P(N|B1)P(B1) + P(N|B2)P(B2) + (P(N|B3)P(B3)}$ = $\frac{(0.297)(0.3)}{(0.297)(0.3) + (0.994)(0.2) + (0.990)(0.5)}$ = 0.1138

Hence the probability that a non-defective product is produced by machine B1 is 11.38%.

4 0

3 years ago