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3 years ago

Need answer Now!! 618 is 75% what is 100%?

Mathematics

2 answers:

aliya0001 [1]3 years ago

6 0

Answer:

824 I think?

Step-by-step explanation:

614 divided by 75 is 8.24.

8.24 x 100 is 824

ch4aika [34]3 years ago

4 0

824, 100% of 824 is 824, therefore 75 percent of 824 equals 618.

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What % of $100 = $90

Harlamova29_29 [7]

Answer: 90%

Step-by-step explanation:

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2 years ago

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Evaluate each expression for t = 0, x = 1.5, y = 6, and z = 23

My name is Ann [436]

Answer:

1. 11

2. 20

3. 34

4. 12.6

5. 18

6. 105

Step-by-step explanation:

We are told that ---> t = 0, x = 1.5, y = 6, and z = 23

1. y + 5

= 6 + 5

= 11

2. z - 2x

= 23 - 2(1.5)

= 23 - 3

= 20

3. 2z - 2y

= 2(23) - 2(6)

= 46 - 12

= 34

4. 2.6y - 2x

= 2.6(6) - 2(1.5)

= 15.6 - 3

= 12.6

5. 4(y - x)

= 4(6 - 1.5)

= 24 - 6

= 18

6. 4(2 + z) + 5

= 4(2 + 23) + 5

= 8 + 92 + 5

= 105

3 0

2 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:

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

Now, calculate the $P (T \geq 5)$ as follows:

$P(T \geq 5)=P(T\geq 5|SP)P(SP)+P(T\geq 5|GS)P(GS)\\=(0.6065)(0.3)+(0.2231)(0.7)=0.3381$

The probability that the train is having signal problems based on the fact that will be at least 5 minutes long is calculated as follows:

$P(SP|T\geq 5)= \frac{0.6065 \times 0.3}{0.3381}
= 0.5381
$

Hence, the probability that the train is having signal problems based on the fact that will be at least 1 minute long is $0.5381$ .

6 0

2 years ago

Alonso went to the market with $55 to buy eggs and sugar. He knows he needs a package of 12 eggs that costs $2.75. After getting

vodka [1.7K]

Answer:

1) The inequality is $55 ≥ $11.50 × S + $2.75

2) Alonso can afford 4 boxes of sugar

Step-by-step explanation:

The given information are;

The total amount with Alonso = $55

The amount of eggs Alonso needs = 12 eggs

The costs of the pack of 12 eggs = $2.75

The number of packs of eggs Alonso buys = 1

The cost of each box of sugar = $11.50

Let the number of boxes of sugar Alonso buys = S

Given that the total amount of money with Alonso is $55, then the inequality will be such that $55 will be equal to or larger than the expression for items bought

Therefore, the inequality is given as follows;

$55 ≥ $11.50 × S + $2.75

2) To find out how many boxes of sugar Alonso can afford, we have;

$55 ≥ $11.50 × S + $2.75

$55 - $2.75≥ $11.50 × S

$52.25 ≥ $11.50 × S

∴ S ≤ $52.25/$11.50 = 4.54

S ≤ 4.54 boxes of sugar

As the sugar comes in whole boxes, Alonso can therefore afford only 4 whole boxes of sugar.

Alonso can afford to buy 4 boxes of sugar.

6 0

2 years ago

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3. Which is equivalent to the expression below?

alexgriva [62]

Answer:

D. 2.2m + 4.5

Step-by-step explanation:

8 0

2 years ago