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Natalka [10]
3 years ago
11

Shelley pours 4/5 gallon of water into a bird bath. The next morning she finds that 3/4 of the water is gone. What is the amount

of water in gallons that is gone?
Mathematics
1 answer:
Sonbull [250]3 years ago
5 0
1/20 a gallon I'm pretty sure. 4/5 gallon is equal to 0.8 minus 3/4 or .75 is equal to 0.05 which is 1/20.
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Campbell’s paycheck was $257.29. She put 1/4 of her paycheck into a savings account and used 1/3 of what was left to pay bills.
Stells [14]

Answer:

She has $128.65 left

Step-by-step explanation:

$257.29 x 1/4 = $64.32

$257.29 - $64.32 = $192.97

$192.97 x 1/3 = $64.32

$192.97 - $64.32 = $128.65

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4 0
3 years ago
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Suppose the area of a circle is 4π. What is its diameter?
mart [117]
Area=pi times radius^2 so

if area=4pi=pi times r^2
4 times pi=r^2 times pi
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radius=1/2 diamters
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6 0
3 years ago
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How to solve this?<br>\int \frac { 4 - 3 x ^ { 2 } } { ( 3 x ^ { 2 } + 4 ) ^ { 2 } } d x​
ivanzaharov [21]

\Large \mathbb{SOLUTION:}

\begin{array}{l} \displaystyle \int \dfrac{4 - 3x^2}{(3x^2 + 4)^2} dx \\ \\ = \displaystyle \int \dfrac{4 - 3x^2}{x^2\left(3x + \dfrac{4}{x}\right)^2} dx \\ \\ = \displaystyle \int \dfrac{\dfrac{4}{x^2} - 3}{\left(3x + \dfrac{4}{x}\right)^2} dx \\ \\ \text{Let }u = 3x + \dfrac{4}{x} \implies du = \left(3 - \dfrac{4}{x^2}\right)\ dx \\ \\ \text{So the integral becomes}  \\ \\ = \displaystyle -\int \dfrac{du}{u^2} \\ \\ = -\dfrac{u^{-2 + 1}}{-2 + 1} + C \\ \\ = \dfrac{1}{u} + C \\ \\ = \dfrac{1}{3x + \dfrac{4}{x}} + C \\ \\ = \boxed{\dfrac{x}{3x^2 + 4} + C}\end{array}

5 0
3 years ago
What is the relationship between the 6s in 660,472​
bixtya [17]
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6 0
3 years ago
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Two equal groups of seedlings, and equal in height, were selected for an experiment. One group of seedlings was fed Fertilizer A
ser-zykov [4K]

Answer:

Null Hypothesis: H_0: \mu_A =\mu _B or \mu_A -\mu _B=0

Alternate Hypothesis: H_1: \mu_A >\mu _B or \mu_A -\mu _B>0

Here to test Fertilizer A height is greater than Fertilizer B

Two Sample T Test:

t=\frac{X_1-X_2}{\sqrt{S_p^2(1/n_1+1/n_2)}}

Where S_p^2=\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}

S_p^2=\frac{(14)0.25^2+(12)0.2^2}{15+13-2}= 0.0521154

t=\frac{12.92-12.63}{\sqrt{0.0521154(1/15+1/13)}}= 3.3524

P value for Test Statistic of P(3.3524,26) = 0.0012

df = n1+n2-2 = 26

Critical value of P : t_{0.025,26}=2.05553

We can conclude that Test statistic is significant. Sufficient evidence to prove that we can Reject Null hypothesis and can say Fertilizer A is greater than Fertilizer B.

6 0
3 years ago
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