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Ksivusya [100]
2 years ago
13

Laura placed a bucket of water in her garden. Over the course of a week, she watched the water evaporate and recorded the volume

of water left in the bucket each day.
Laura found the linear model that best fit the data was V=5.00−0.25n, where n is the number of days since she first placed the bucket and V is the volume of water, in liters, remaining in the bucket.

How many liters of water evaporated from the bucket every day?
How may liters where inside the bucket when Laura first placed it in the garden?
Mathematics
1 answer:
nikdorinn [45]2 years ago
4 0

Answer:

1. 0.5 L; 2. 5.00 L  

Step-by-step explanation:

V = 5.00 - 0.5n

If you include units, the equation becomes

V(in litres) = 5.00 L - (0.5 L/day) × (n days)

1. Rate of evaporation

When you include the units, it becomes easier to see that the water is evaporating at a rate of 0.5 L/day.

That is, 0.5 L of water evaporates each day.

The negative sign shows that the volume of water is decreasing.

2. Volume at the beginning

At the beginning of the experiment, n = 0. Then

V = 5.00 -0.5×0 = 5.00 - 0 = 5.00 L

The bucket originally contained 5.00 L of water.

 

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10. If PQRS is a parallelogram, then P - R is:<br>(a) 20<br>(b) 30<br>(c) 90<br>(d) 0​
statuscvo [17]

Answer:

d) 0

Step-by-step explanation:

P -R = 0

Because in parallelogram, opposite angles are equal.

4 0
2 years ago
A study of 31,000 hospital admissions in New York State found that 4% of the admissions led to treatment-caused injuries. One-se
vovikov84 [41]

Answer:

Step-by-step explanation:

a) probability a person admitted to the hospital will suffer a treatment-caused injury due to negligence

P(injury) = 4%

P(negligence) = 1/4 = 0.25

We need to find probability (injury)(negligence)

P(injury) * P(negligence) = 0.04*0.25 = 0.01

b) probability a person admitted to the hospital will die from a treatment-caused injury

P(injury) = 4%

P(death) = 1/7

P(Injury) *P(death) = 0.04/7 = 0.00571

c) In the case of a negligent treatment-caused injury, what is the probability a malpractice claim will be paid

P(claim) = 1/7.5

P(payment) = 1/2

P(claim)*P(payment) = 1/7.5 * 1/2 = 0.06

7 0
2 years ago
Based on the graph shown to the right, whose phones cost varies directly with the amount of time spent talking?
Lena [83]

Answer:

B

Step-by-step explanation:

Given 2 quantities that vary directly, then the graph must pass through the origin.

Nikiya's graph is the only one to do this ⇒ B



8 0
2 years ago
Read 2 more answers
Solve the system of equations – 5x + y = 56 and x + y = -4 by combining the<br> equations.
anygoal [31]

Answer:

Your solution is (-10, 6).

Step-by-step explanation:

Combining the equations is also known as substitution. This is done when you substitute one variable into another equation.

-5x + y = 56

x + y = -4

Let's change the second equation into one with one variable on each side.

y = -x - 4

Now, plug this into your first equation.

-5x + (-x - 4) = 56

Distribute the + sign.

-5x - x - 4 = 56

Combine the like terms.

-6x - 4 = 56

-6x = 60

Isolate x by dividing both sides by -6.

x = -10

Now plug this back into either equation.

-10 + y = -4

Add 10 to both sides to find y.

y = 6

Your solution is (-10, 6).

Check this by plugging in these values into the equation you have not checked yet.

-5(-10) + (6) = 56

50 + 6 = 56

56 = 56

Your solution is correct.

Hope this helps!

5 0
1 year ago
Read 2 more answers
Suppose that the IQs of university​ A's students can be described by a normal model with mean 140 and standard deviation 8 poi
AnnZ [28]

Answer:

0.0266, 0.9997,0.7856

Step-by-step explanation:

Given that the IQs of university​ A's students can be described by a normal model with mean 140 and standard deviation 8 points. Also suppose that IQs of students from university B can be described by a normal model with mean 120 and standard deviation 11. Let x be the score by A students and Y the score of B.

A)P(X>135) = \\P(Z>0.625)\\=0.0266

B) Since X and Y are independent we have

X-Y is Normal with mean = 140-120 =20 and Var (x-y)=Var(x)+Var(y) = 19

P(X-Y)>5\\\\=1-0.00029\\=0.9997

C) For a group of 3, average has std deviation = \frac{11}{\sqrt{3} } \\=6.351

P(\bar y >115)\\= P(z>\frac{-5}{6.351} \\=0.7856

3 0
2 years ago
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