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2 years ago

What is the rate of change: y=x+5

Mathematics

1 answer:

const2013 [10]2 years ago

4 0

Answer:

x or 1x.

Step-by-step explanation:

The rate of change is slope.

Slope=m

y=mx+b

y=<u>x</u>+5

x is the same thing as 1x. So...

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What percent of 150 is 9?

irga5000 [103]

One of the ways to find the answer is to find how much 1 percent of 150 is. To do that we just have to divide 150 by 100:

150 / 100 = 1.5

Now we can divide 9 by 1.5 to know how much percent of 150 it is:

9 / 1.5 = 6

Let's check it:

150 * 6% = 150 * 6/100 = 15 * 6/10 = 3 * 6/2 = 18/2 = 9

It's correct then :)

3 0

3 years ago

Question: solve for x. <br><br>please someone help!

Dominik [7]

by angle sum property of triangle,

70+60+(x+60)=180

=> 190 +x = 180

=> x = 180-190

=> x = -10

5 0

3 years ago

Two polygons are similar. The perimeter of the smaller polygon is 66 feet and the ratio of the corresponding side lengths is 3 4

Alisiya [41]

Answer:

Perimeter of the other polygon is 88 feet.

Step-by-step explanation:

We are given that, 'Two polygons are similar'.

Also, the perimeter of the smaller polygon is 66 feet and the ratio of the corresponding side lengths is $\frac{3}{4}$ .

As, we know,

'If two polygons are similar, the ratio of their perimeters is equal to the ratio of their side lengths'.

Thus, we have,

Let, the perimeter of the other polygon = x feet.

$\frac{3}{4}=\frac{66}{x}$

i.e. $x=\frac{66\times 4}{3}$

i.e. $x=\frac{264}{3}$

i.e. x= 88 feet

Thus, the perimeter of the other polygon is 88 feet.

8 0

3 years ago

A subway has good service 70% of the time and runs less frequently 30% of the time because of signal problems. When there are si

nordsb [41]

Answer:

Step-by-step explanation:

Let GS denote the good service and SP denote the signal problem.

A subway has good service 70% of the time, that is, $P(GS)=0.7$ and a subway runs less frequently 30% of the time because of the signal problems, that is, $P(SP)=0.3$ .

If there are signal problems, the amount of time T in minutes that have to wait at the platform is described by the probability density function given below:

$P_{T|SP}(t)=0.1e^{0.1t}$

If there is good service, the amount of time T in minutes that have to wait at the platform is described the probability density function given below:

$P_{T|GOOD}(t)=0.3e^{0.3t}$

(a)

The probability that you wait at least 1 minute if there is good service P(T ≥ 1| GS) is obtained as follows:

$P(T\geq 1|GS)=\int\limits^{\infty}_1 {0.3e^{-0.3t}} \, dt\\\\=0.3\int\limits^{\infty}_1 {e^{-0.3t}} \, dt\\\\=0.3[(\frac{e^{-0.3t}}{-0.3})]\\\\=-(e^{-0.3t})\limits^{\infty}_1\\\\=-(0-e^{-0.3})\\=0.74$

(b)

The probability that you wait at least 1 minute if there is signal problems P(T ≥ 1| SP) is obtained as follows:

$P(T\geq 1|SP)=\int\limits^{\infty}_1 {0.1e^{-0.1t}} \, dt\\\\=0.1\int\limits^{\infty}_1 {e^{-0.1t}} \, dt\\\\=0.1[(\frac{e^{-0.1t}}{-0.3})]\\\\=-(e^{-0.1t})\limits^{\infty}_1\\\\=-(e^{\infty}-e^{-0.1})\\=-(0-0.904)\\=0.904$

(c)

After 1 minute of waiting on the platform, the train is having signal problems follows an

exponential distribution with parameter $\lambda= 0.1$

The probability that the train is having signal problems based on the fact that will be at least 1 minute long is obtained using the result given below:

$P(SP|T\geq 1)=\frac{P(T\geq 1|SP)P(SP)}{P(T\geq 1)}$

$P(T\geq 1|GS)=0.74, P(T\geq 1|SP)=0.904$

Now calculate the $P(T \geq 1)$ as follows:

$P(T \geq 1)=P(T\geq 1|SP)P(SP)+P(T\geq 1|GS)P(GS)\\=(0.904)(0.3)+(0.74)(0.7)=0.7892$

The probability that the train is having signal problems based on the fact that will be at least 1 minute long is calculated as follows:

$P(SP|T\geq 1)= \frac{0.904 \times 0.3}{0.7892}
= 0.3436
$

Hence, the probability that the train is having signal problems based on the fact that will be at least 1 minute long is $0.3436$ .

(d)

After 5 minutes of waiting on the platform, the train is having signal problems follows an exponential distribution with parameter $\lambda= 0.1$ .

The probability that the train is having signal problems based on the fact that will be at least 5 minutes long is obtained using the result given below:

$P(SP|T\geq 5)=\frac{P(T\geq 5|SP)P(SP)}{P(T\geq 5)}$

First, calculate the $P(T\geq 5|SP)$ as follows:

$P(T\geq 5|SP)=\int\limits^{\infty}_5 {0.1e^{-0.1t}} \, dt\\\\=0.1\int\limits^{\infty}_5 {e^{-0.1t}} \, dt\\\\=0.1[(\frac{e^{-0.1t}}{-0.1})]\\\\=-(e^{-0.1t})\limits^{\infty}_5\\\\=-(e^{\infty}-e^{-0.5})\\=-(0-0.6065)\\=0.6065$

Now, calculate the $P (T\geq5|GS )$ as follows:

$P(T\geq 5|GS)=\int\limits^{\infty}_5 {0.3e^{-0.3t}} \, dt\\\\=0.3\int\limits^{\infty}_5 {e^{-0.3t}} \, dt\\\\=0.3[(\frac{e^{-0.3t}}{-0.3})]\\\\=-(e^{-0.3t})\limits^{\infty}_5\\\\=-(0-e^{-1.5})\\=0.2231$