All categories

3 years ago

How many solutions do linear inequalities and linear equations have each?

1 answer:

6 0

With linear equations, there are only three possibilities: There are zero solutions. There is one solution. There are infinitely many solutions.

You might be interested in

Can someone please help me figure out how to do this problem. We have tried several different ways and none of them come up as b

Ilya [14]

Answer:

Step-by-step explanation:

Order of operations: PEMDAS (parentheses, exponents, multiplication, division, addition, subtraction)

5(9)+2(6) ÷ 3 + $5^{2}$

= 45+12÷ 3 +25

=45+4+25

=74

8 0

3 years ago

Read 2 more answers

Right answer will get a brainlist

Vladimir [108]

It is 12.31 I think.

5 0

3 years ago

A ball is launched into the sky at 272 feet per second from the roof of a skyscraper 1,344 feet tall. The equation for the ball’

xz_007 [3.2K]

When the ball is in the ground, its height is basically equal to zero thus making our equation,

-16t^2 + 272t + 1344 = 0

Simplifying the equation will give us,

- t^2 + 17t + 84 = 0 or t^2 – 17t – 84 = 0

Factoring out the equation will give us,

(t – 21)(t + 4) = 0

Thus, t = 21 or t = -4. -4 is an extraneous root. Thus, the answer is t = 21.

Answer: 21 seconds

5 0

3 years ago

A person traveling from Seattle to Sydney has three airlines to choose from. 40% of travelers choose airline A, and this airline

dalvyx [7]

Answer:

46.67% probability that they flew with airline B.

Step-by-step explanation:

We have these following probabilities:

A 40% probability that a traveler chooses airline A.

A 35% probability that a traveled chooses airline B.

A 25% probability that a traveler chooses airline C.

If a passenger chooses airline A, a 10% probability that he arrives late.

If a passenger chooses airline B, a 15% probability that he arrives late.

If a passenger chooses airline C, a 8% probability that he arrives late.

If a randomly selected traveler is on a flight from Seattle which arrives late to Sydney, what is the probability that they flew with airline B?

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

What is the probability that the traveler flew with airline B, given that he was late?

P(B) is the probability that he flew with airline B.

So $P(B) = 0.35$

P(A/B) is the probability of being late when traveling with airline B. So P(A/B) = 0.15.

P(A) is the probability of being late. This is the sum of 10% of 40%(airline A), 15% of 35%(airline B) and 8% of 25%(airline C).

$P(A) = 0.1*0.4 + 0.15*0.35 + 0.08*0.25 = 0.1125$

Probability

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.35*0.15}{0.1125} = 0.4667$

46.67% probability that they flew with airline B.

5 0

3 years ago

Meredith invested $5,500 into an account that earned 5% interest per year. When she cashed out the investment to use it as a dow

Andrej [43]

Answer: B. $7739.05=5500(1.05)^t$

Step-by-step explanation:

The formula to find the compound amount (compounded yearly):-

$A=P(1+r)^t$ , where P is the principal amount invested, is the rate of interest and t is time period.

As per given , we have

P=$5500 , r=5% = 0.05 and A = $7,739.05.

Substitute all the values in the formula , we get

$7739.05=5500(1+0.05)^t$

$7739.05=5500(1.05)^t$

Hence, the equation describes Meredith's investment based on t, the number of years she kept the account open : $7739.05=5500(1.05)^t$

5 0

3 years ago