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2 years ago

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OABC is a tetrahedron and OA=a, OB=b, and OC=c. The point P and Q are such that OA =AP and 2OB = BQ. The point M is a midpoint o

f PQ. Find (i)AB (ii)PQ (iii)CQ (iv)QM (v)MB (vi)OM in terms of a, b, and c.

1 answer:

serg [7]2 years ago

5 0

Solution:

Let OA=a, OB=b, and OC=c

Then OP=2a and OQ=3b

OA+AB=OB ⇒ AB=b-a

OP+PQ=OQ ⇒ PQ = 3b-2a

OC+CQ = OQ ⇒ CQ = 3b-c

$QM= -\frac{1}{2} PQ = a-\frac{3}{2}b$

$OM=OQ+QM = 3b + a-\frac{3}{2}b = a+\frac{3}{2}b$

$OM+MB=OB=>MB = b-(a+\frac{3}{2}b) = -a-\frac{1}{2}b$

(i) AB = b-a

(ii) PQ = 3b-2a

(iii) CQ = 3b-c

(iv) QM = $a-\frac{3}{2}b$

(v) MB = $-a-\frac{1}{2}b$

(vi) OM = $a+\frac{3}{2}b$

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Which of the following are dimensionally consistent? (Choose all that apply.)(a) a=v / t+xv2 / 2(b) x=3vt(c) xa2=x2v / t4(d) x=v

Bumek [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

A

is dimensionally consistent

B

is not dimensionally consistent

C

is dimensionally consistent

D

is not dimensionally consistent

E

is not dimensionally consistent

F

is dimensionally consistent

G

is dimensionally consistent

H

is not dimensionally consistent

Step-by-step explanation:

From the question we are told that

The equation are

$A) \ \ a^3 = \frac{x^2 v}{t^5}$

$B) \ \ x = t$

$C \ \ \ v = \frac{x^2}{at^3}$

$D \ \ \ xa^2 = \frac{x^2v}{t^4}$

$E \ \ \ x = vt+ \frac{vt^2}{2}$

$F \ \ \ x = 3vt$

$G \ \ \ v = 5at$

$H \ \ \ a = \frac{v}{t} + \frac{xv^2}{2}$

Generally in dimension

x - length is represented as L

t - time is represented as T

m = mass is represented as M

Considering A

$a^3 = (\frac{L}{T^2} )^3 = L^3\cdot T^{-6}$

and $\frac{x^2v}{t^5 } = \frac{L^2 L T^{-1}}{T^5} = L^3 \cdot T^{-6}$

Hence

$a^3 = \frac{x^2 v}{t^5}$ is dimensionally consistent

Considering B

$x = L$

and

$t = T$

Hence

$x = t$ is not dimensionally consistent

Considering C

$v = LT^{-1}$

and

$\frac{x^2 }{at^3} = \frac{L^2}{LT^{-2} T^{3}} = LT^{-1}$

Hence

$v = \frac{x^2}{at^3}$ is dimensionally consistent

Considering D

$xa^2 = L(LT^{-2})^2 = L^3T^{-4}$

and

$\frac{x^2v}{t^4} = \frac{L^2(LT^{-1})}{ T^5} = L^3 T^{-5}$

Hence

$xa^2 = \frac{x^2v}{t^4}$ is not dimensionally consistent

Considering E

$x = L$

;

$vt = LT^{-1} T = L$

and

$\frac{vt^2}{2} = LT^{-1}T^{2} = LT$

Hence

$E \ \ \ x = vt+ \frac{vt^2}{2}$ is not dimensionally consistent

Considering F

$x = L$

and

$3vt = LT^{-1}T = L$ Note in dimensional analysis numbers are

not considered

Hence

$F \ \ \ x = 3vt$ is dimensionally consistent

Considering G

$v = LT^{-1}$

and

$at = LT^{-2}T = LT^{-1}$

Hence

$G \ \ \ v = 5at$ is dimensionally consistent

Considering H

$a = LT^{-2}$

,

$\frac{v}{t} = \frac{LT^{-1}}{T} = LT^{-2}$

and

$\frac{xv^2}{2} = L(LT^{-1})^2 = L^3T^{-2}$

Hence

$H \ \ \ a = \frac{v}{t} + \frac{xv^2}{2}$ is not dimensionally consistent

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2 years ago

The price of a math book after a discount of 25% is $36. What is the original price of the math book?

UNO [17]

If $36=75%,
That means we can divide both sides by three to get $12=25%,
And because 25x4 is 100, we multiply both sides of the equation by 4 to get $48=100%,
So the answer is $48 dollars.

This also worked for me but I’m not sure it’s the most reliable way:
36 ÷ 0.75 = 48
(amount after discount ÷ percentage that is remaining)

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2 years ago

Elena wants to make a scale drawing of her bedroom. Her bedroom is a rectangle with length 5 m and width 3 m. She decides on a s

katovenus [111]

We are given dimensions of rectangular bedroom = length of 5m and width 3 m.

Also scale drawing of factor 1/50 of the original dimension.

In order to find the dimensions of scale drawing, we need to multiply scale factor by each dimension.

Therefore, length of the scale drawing is 5 * 1/50 = 1/10 m and

Let us convert it in centimeter now.

1 m = 100 cm

1/10 m = 100 * 1/10 = 10cm.

Width of the scale drawing = 3 * 1/50 = 3/50 m.

100*3/50 = 6 cm.

Therefore, the dimensions of a scale drawing of Elenas bedroom is length 10cm and width 6cm.

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What dose 0.07x1.22 eaqual

iogann1982 [59]

0.07 * 1.22

Is equal to 0.0854

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2 years ago

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g Assume that the distribution of time spent on leisure activities by adults living in household with no young children is norma

OLga [1]

Answer:

"<em>The probability that the amount of time spent on leisure activities per day for a randomly selected adult from the population of interest is less than 6 hours per day</em>" is about 0.8749.

Step-by-step explanation:

We have here a <em>random variable</em> that is <em>normally distributed</em>, namely, <em>the</em> <em>time spent on leisure activities by adults living in a household with no young children</em>.

The normal distribution is determined by <em>two parameters</em>: <em>the population mean,</em> $\\ \mu$ , and <em>the population standard deviation,</em> $\\ \sigma$ . In this case, the variable follows a normal distribution with parameters $\\ \mu = 4.5$ hours per day and $\\ \sigma = 1.3$ hours per day.

We can solve this question following the next strategy:

Use the <em>cumulative</em> <em>standard normal distribution</em> to find the probability.
Find the <em>z-score</em> for the <em>raw score</em> given in the question, that is, <em>x</em> = 6 hours per day.
With the <em>z-score </em>at hand, we can find this probability using a table with the values for the <em>cumulative standard normal distribution</em>. This table is called the <em>standard normal table</em>, and it is available on the Internet or in any Statistics books. Of course, we can also find these probabilities using statistics software or spreadsheets.

We use the <em>standard normal distribution </em>because we can "transform" any raw score into <em>standardized values</em>, which represent distances from the population mean in standard deviations units, where a <em>positive value</em> indicates that the value is <em>above</em> the mean and a <em>negative value</em> that the value is <em>below</em> it. A <em>standard normal distribution</em> has $\\ \mu = 0$ and $\\ \sigma = 1$ .

The formula for the <em>z-scores</em> is as follows

$\\ z = \frac{x - \mu}{\sigma}$ [1]

Solving the question

Using all the previous information and using formula [1], we have

<em>x</em> = 6 hours per day (the raw score).

$\\ \mu = 4.5$ hours per day.

$\\ \sigma = 1.3$ hours per day.

Then (without using units)

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

$\\ z = \frac{6 - 4.5}{1.3}$

$\\ z = \frac{1.5}{1.3}$

$\\ z = 1.15384 \approx 1.15$

We round the value of <em>z</em> to two decimals since most standard normal tables only have two decimals for z.

We can observe that z = 1.15, and it tells us that the value is 1.15 standard deviations units above the mean.

With this value for <em>z</em>, we can consult the <em>cumulative standard normal table</em>, and for this z = 1.15, we have a cumulative probability of 0.8749. That is, this table gives us P(z<1.15).

We can describe the procedure of finding this probability in the next way: At the left of the table, we have z = 1.1; we can follow the first line on the table until we find 0.05. With these two values, we can determine the probability obtained above, P(z<1.15) = 0.8749.

Notice that the probability for the z-score, P(z<1.15), of the raw score, P(x<6) are practically the same, $\\ P(z$ . For an exact probability, we have to use a z-score = 1.15384 (without rounding), that is, $\\ P(z$ . However, the probability is approximated since we have to round z = 1.15384 to z = 1.15 because of the use of the table.

Therefore, "<em>the probability that the amount of time spent on leisure activities per day for a randomly selected adult from the population of interest is less than 6 hours per day</em>" is about 0.8749.

We can see this result in the graphs below. First, for P(x<6) in $\\ N(4.5, 1.3)$ (red area), and second, using the standard normal distribution ( $\\ N(0, 1)$ ), for P(z<1.15), which corresponds with the blue shaded area.

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