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2 years ago

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A dog's weight increased by 50% in 3 years. by the end of 3 years, the dog weighed 45 pounds. how much did the dog weigh 3 years

ago?

2 answers:

weeeeeb [17]2 years ago

6 0

I hope this helps you

45+45.50%

45+45.50/100

45+22,5

67,5

bogdanovich [222]2 years ago

4 0

Answer:

<h2><em><u>30</u></em></h2>

Step-by-step explanation:

<h2><u><em>x * 1.5 = 45 </em></u></h2><h2><u><em>x = 45/1.5 </em></u></h2><h2><u><em>x = 30</em></u></h2>

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In order to construct this equation, we will use the variables:
V to represent mixture volume (40 ml)
C to represent mixture concentration (0.32)
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v₂ to represent the volume of the second solution (40 * 3/4 = 30 ml)
c₂ to represent the concentration of the second solution

We know that the total amount of substance, product of the volume and concentration, in the final solution is equal to the individual amounts in the two given solutions. Thus:
VC = v₁c₁ + v₂c₂
40(0.32) = 10(0.2) + 30c

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2 years ago

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Sam is observing the velocity of a car at different times. After two hours, the velocity of the car is 50 km/h. After six hours,

Eva8 [605]

Let:
x = hours of travel
y = velocity

slope= rise/run slope=(y2-y1)/(x2-x1)
(x1,y1) = (2,50) (x2,y2) = (6,54)
sub values back into the equation m = (54-50)/(6-2) m = 1
POINT SLOPE FORMy-y1 = m(x-x1) y-50= 1(x-2) y = x -2 +50
y = x + 48

B)

the graph within the first seven hours can be obtained at point B

x = 7

y = 7+48 = 55
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Josef owns four par value $1,000 bonds from Dowc Beverage Co. Each bond has a market value of 104.561 and gives 9.2% interest. J

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the time taken by a student to the university has been shown to be normally distributed with mean of 16 minutes and standard dev

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Answer:

a) 2.84% probability that he is late for his first lecture.

b) 5.112 days

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 16, \sigma = 2.1$

a. Find the probability that he is late for his first lecture.

This is the probability that he takes more than 20 minutes to walk, which is 1 subtracted by the pvalue of Z when X = 20. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{20 - 16}{2.1}$

$Z = 1.905$

$Z = 1.905$ has a pvalue of 0.9716

1 - 0.9716 = 0.0284

2.84% probability that he is late for his first lecture.

b. Find the number of days per year he is likely to be late for his first lecture.

Each day, 2.84% probability that he is late for his first lecture.

Out of 180

0.0284*180 = 5.112 days

4 0

1 year ago

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coldgirl [10]

Answer:

The minimum cost would be 480$ when Inna works for 8 hours and Jim works for 20 hours.

Step-by-step explanation:

We are given the following information in the question:

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Number of pages typed by Inna in 1 hour = 6

Charges for 1 hour for Jim = $18

Number of pages typed by Jim in 1 hour = 8

Let x be the number of hours Inna work and let y be the number of hours Jim work.

Total cost = $15x + 18y$

We have to minimize this cost.

Then, we can write the following inequalities:

$6x + 8y \geq 208\\x \geq 8\\y \geq 8\\$

The corner points as evaluated from graph are: (8,20) and (24,8)

C(8,20) = 480$

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Hence, the minimum cost would be 480$ when Inna works for 8 hours and Jim works for 20 hours.

The attached image shows the graph.

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1 year ago