All categories

3 years ago

The tv costs $850 you pay 20% , how much is left to pay

Mathematics

2 answers:

WINSTONCH [101]3 years ago

7 0

850 x .2 = 170 So you still have to pay $170

kogti [31]3 years ago

3 0

The answer would be 680 because 20% of 850 is 170 so 850-170=680.

You might be interested in

The company that owns the telescope gives a discount of $8 off the rental fee for every rental hour after the first two hours. T

NISA [10]

Answer:

Step-by-step explanation:Please give me a brainly

4 0

3 years ago

In the following image, AB is parallel to DC, and BC is a transversal intersecting both parallel lines.

Ulleksa [173]

Answer:

In the following image, AB is parallel to DC, and BC is a transversal intersecting both parallel lines.

and the angle marked q° are ________angles.

The angles marked p° and q° are ________

The angle marked p° and are _______angles.

The angles marked q° and n° are _______angles.

The angles marked n° and m° are _______angles.

DONT ANSWER UNLESS YOU KNOW IT

6 0

3 years ago

WILL GIVE BRAINLIESTTT!!!! PLS HELP

Kaylis [27]

7/30 x 100 = approx 23.3%

4 0

3 years ago

What number goes in the highlighted box?

pentagon [3]

-15+7=-8
Adding numbers with different signs makes it subtraction and the answer will have the sign of the bigger number

3 0

3 years ago

A machine has four components, A, B, C, and D, set up in such a manner that all four parts must work for the machine to work pro

olchik [2.2K]

Answer:

0.756

Step-by-step explanation:

It is given that a machine has four components, A, B, C, and D.

$P(A)=P(B)=0.93, P(C)=0.95,P(D)=0.92$

If these components set up in such a manner that all four parts must work for the machine to work properly.

We need to find the probability that the machine works properly. It means we have to find the value of $P(A\cap B\cap C\cap D)$ .

If two events X and Y are independent, then

$P(X\cap Y)=P(X)\times P(Y)$

Assume the probability of one part working does not depend on the functionality of any of the other parts.

$P(A\cap B\cap C\cap D)=P(A)\times P(B)\times P(C)\times P(D)$

Substitute the given values.

$P(A\cap B\cap C\cap D)=0.93\times 0.93\times 0.95\times 0.92$

$P(A\cap B\cap C\cap D)=0.7559226$

$P(A\cap B\cap C\cap D)\approx 0.756$

Therefore, the probability that the machine works properly is 0.756.

7 0

3 years ago