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Elanso [62]
3 years ago
5

2.1(0.2x−1.4)=1.3(0.4x−3.5) simplify lots of points

Mathematics
1 answer:
Zina [86]3 years ago
5 0

Answer:

you need a y to get at least 1 point

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A delivery truck is transporting boxes of two sizes: large and small. The large boxes weigh 55 pounds each, and the small boxes
nikdorinn [45]

Answer:

The number of large boxes is 55 and the number of small boxes is 70

Step-by-step explanation:

<u><em>The complete question is</em></u>

A delivery truck is transporting boxes of two sizes: large and small. The large boxes weigh 50 pounds each, and the small boxes weigh 20 pounds each. There are 125 boxes in all. If the truck is carrying a total of 4150 pounds in boxes, how many of each type of box is it carrying?

Let

x ------> the number of large boxes

y -----> the number of small boxes

we know that

x+y=125 -----> equation A

50x+20y=4,150 ----> equation B

Solve the system by graphing

Remember that the solution of the system of equations is the intersection point both graphs

The intersection point is (55,70)

therefore

The number of large boxes is 55 and the number of small boxes is 70

3 0
2 years ago
Apollo traveled 84 leagues in 6 days. Then he doubled his rate. How long would it take him
LenKa [72]
168 leagues in 6 days
4 0
3 years ago
Read 2 more answers
This year, a herd of bison had 10% increase in population. If there was 550 bison in the herd last year, how many are in the her
sattari [20]

Answer: 605

Step-by-step explanation:

Last year            This year

550 x 1.1      =     605

7 0
3 years ago
Read 2 more answers
A hospital’s financial reimbursement payor mix indicates 35% of the patients have Medicare and 15% have Medicaid. Of those who h
Damm [24]

Using conditional probability, it is found that there is a 0.035 = 3.5% probability that a hospital patient has both Medicare and Medicaid.

<h3>What is Conditional Probability?</h3>
  • <em>Conditional probability</em> is the <u>probability of one event happening, considering a previous event</u>. The formula is:

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

  • P(B|A) is the probability of event B happening, given that A happened.
  • P(A \cap B) is the probability of both A and B happening.
  • P(A) is the probability of A happening.

In this problem, the events are:

  • Event A: Patient has Medicare.
  • Event B: Patient has Medicaid.

For the probabilities, we have that:

  • 35% of the patients have Medicare, hence P(A) = 0.35.
  • Of those who have Medicare, there is a 10% chance they also have Medicaid, hence P(B|A) = 0.1.

Then, applying the <em>conditional </em>probability:

P(B|A) = \frac{P(A \cap B)}{P(A)}

0.1 = \frac{P(A \cap B)}{0.35}

P(A \cap B) = 0.35(0.1)

P(A \cap B) = 0.035

0.035 = 3.5% probability that a hospital patient has both Medicare and Medicaid.

You can learn more about conditional probability at brainly.com/question/14398287

4 0
2 years ago
Help me please ;-; ;-; ;-; ;-;
Evgesh-ka [11]

Answer:

<em>d</em>= 4/7 or 0.571428



7 0
3 years ago
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