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3 years ago

What is the condition of loss ?1)sp>mp 2)CP>sp 3) sp>CP 4)mp>sp

Mathematics

1 answer:

victus00 [196]3 years ago

3 0

Step-by-step explanation:

what is the condition of loss ?1)sp>mp 2)CP>sp 3) sp>CP 4)mp>sp

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The perimeter of a rectangle is about 13.9 cm. If you double the width of the rectangle the perimeter is about 17.8 cm. Find the

fredd [130]

Answer:

5cm by 1.95cm

Step-by-step explanation:

Let the length of the rectangle be x

Let the width be y

Perimeter of a rectangle = 2x + 2y

If the initial perimeter of a rectangle is about 13.9 cm, then;

13.9 = 2x + 2y ..... 1

If when the width is double the perimeter is 17.8cm, then;

17.8 = 2x + 4y ..... 2

Subtract 1 from 2;

13.9 - 17.8 = 2y - 4y

- 3.9 = -2y

y = 3.9/2

y = 1.95 cm

Substitute y = 1.95 into 1 to get x;

From 1;

13.9 = 2x + 2y

13.9 = 2x + 2(1.95)

13.9 = 2x + 3.9

2x = 13.9-3.9

2x = 10

x = 5 cm

Hence the dimension of the smaller triangle is 5cm by 1.95cm

7 0

3 years ago

A student S is suspected of cheating on exam, due to evidence E of cheating being present. Suppose that in the case of a cheatin

Kay [80]

Answer:

Step-by-step explanation:

Suppose we think of an alphabet X to be the Event of the evidence.

Also, if Y be the Event of cheating; &

Y' be the Event of not involved in cheating

From the given information:

$P(\dfrac{X}{Y}) = 60\% = 0.6$

$P(\dfrac{X}{Y'}) = 0.01\% = 0.0001$

$P(Y) = 0.01$

Thus, $P(Y') \ will\ be = 1 - P(Y)$

P(Y') = 1 - 0.01

P(Y') = 0.99

The probability of cheating & the evidence is present is = P(YX)

$P(YX) = P(\dfrac{X}{Y}) \ P(Y)$

$P(YX) =0.6 \times 0.01$

$P(YX) =0.006$

The probabilities of not involved in cheating & the evidence are present is:

$P(Y'X) = P(Y') \times P(\dfrac{X}{Y'})$

$P(Y'X) = 0.99 \times 0.0001 \\ \\ P(Y'X) = 0.000099$

(b)

The required probability that the evidence is present is:

P(YX or Y'X) = 0.006 + 0.000099

P(YX or Y'X) = 0.006099

(c)

The required probability that (S) cheat provided the evidence being present is:

Using Bayes Theorem

$P(\dfrac{Y}{X}) = \dfrac{P(YX)}{P(Y)}$

$P(\dfrac{Y}{X}) = \dfrac{P(0.006)}{P(0.006099)}$

$P(\dfrac{Y}{X}) = 0.9838$

5 0

3 years ago

A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks along the x-axis from the spotlight toward the bui

Sedbober [7]

Answer:

0.675 m/s

Step-by-step explanation:

Let height of shadow= y,CD=x

Height of man=2 m

Speed of man= $\frac{dx}{dt}=1. 8 m/s$

$\triangle ABD\sim\triangle ECD$

Therefore, $\frac{AB}{EC}=\frac{BD}{CD}$

$\frac{y}{2}=\frac{12}{x}$

$xy=24$

Differentiate w.r.t t

$x\frac{dy}{dt}+y\frac{dx}{dt}=0$

$x\frac{dy}{dt}=-y\frac{dx}{dt}$

$\frac{dy}{dt}=-\frac{y}{x}\frac{dx}{dt}$

When the man is 4 m from the building

Then, we have x=12-4=8 m

$\frac{dx}{dt}=1.8 m/s$

Substitute the values in above equation then, we get

$8y=24$

$y=\frac{24}{8}=3$

Substitute the values then we get

$\frac{dy}{dt}=-\frac{3}{8}\times 1.8=-0.675 m/s$

Hence, the length of his shadow on the building decreasing at the rate 0.675 m/s.

8 0

3 years ago

Rectangle ABCD is graphed in the coordinate plane. The following are the vertices of the rectangle: A(2, -6), B(5, -6), C(5, -2)

max2010maxim [7]

Answer:

Step-by-step explanation:

Refer the answer of brainly.com/question/17115212

3 0

3 years ago

How many 2s are 120

sammy [17]

Answer: 2 can go into 120 60 times.

Step-by-step explanation:

120/2 = 60

6 0

3 years ago

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What is the condition of loss ?1)sp>mp 2)CP>sp 3) sp>CP 4)mp>sp​

What is the condition of loss ?1)sp>mp 2)CP>sp 3) sp>CP 4)mp>sp