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irga5000 [103]
3 years ago
15

Today only, a table is being sold for $333 . This is 74% of its regular price. What was the price yesterday?

Mathematics
1 answer:
Vinil7 [7]3 years ago
5 0
The table was being sold at $450 yesterday
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The height of a building is 1734 ft. How long would it take an object to fall to the ground from the​ top? Use the formula s equ
koban [17]

Answer:

t is approximately 10.41 seconds

Step-by-step explanation:

s = 16t^2

1734 = 16t^2

divide by 16

1734/16 = 16t^2/16

1734/16 = t^2

take the square root  of each side

sqrt(1734/16) = sqrt(t^2)

we only take the positive square root because time must be positive

sqrt(1734/16) = t

t is approximately 10.41 seconds

7 0
3 years ago
Solve 3(x - 2) &lt; 18.<br> Please help me
lakkis [162]

Answer:

x > 8

Step-by-step explanation:

I hope this helps you out!

4 0
3 years ago
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Dairy farmers are aware there is often a linear relationship between the age, in years, of a dairy cow and the amount of milk pr
Jobisdone [24]

Answer:

B. A cow of 5 years is predicted to produce 5.5 more gallons per week.

Step-by-step explanation:

Let M(a) = 40.8-1.1\cdot a, where a is the age of the dairy cow, measured in years, and M(a) is the predicted milk production, measured in gallons per week.

Besides, we consider a_{1} and a_{2}, such that a_{1}\ne a_{2}, we define the difference between predicted milk productions (\Delta M) below:

\Delta M = -1.1\cdot (a_{2}-a_{1}) (1)

If we know that a_{1} = 5\,yr and a_{2} = 10\,yr, then the difference between predicted milk productions is:

\Delta M = -1.1\cdot (10-5)

\Delta M = -5.5\,\frac{gal}{week}

That is, a cow of 5 years is predicted to produce 5.5 more gallons per week than a cow of 10 years. Hence, the right answer is B.

6 0
3 years ago
Save and
Darya [45]

Answer: 6/5

Step-by-step explanation:

7 0
3 years ago
During the grand opening of the T-Shirt Shack at the Madera County Emu Farm, four of the emus broke through the fence and attack
Irina-Kira [14]

Given :

p = 0.68

margin of error, E = 0.1

99% confident interval

alpha, α = 1 - 0.99

             = 0.01

$\frac{\alpha }{2}=\frac{0.01}{2} = 0.005$

∴ $z_{\alpha/2}=z_{0.005}=2.5758$ (z-critical value)

Sample size = $z_{\alpha/2} \times \frac{p(1-p)}{E^2}$

                    $=(2.5758)^2 \times 0.68 \times \frac{(1-0.68)}{0.1^2}$

                    $= 6.63475 \times 0.68 \times \frac{0.32}{0.01}$

                   = 144.37

                  = 145 (rounded off)

Therefore, there are 145 businesses does she need to sample to meet this criteria.

6 0
3 years ago
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