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Cloud [144]
3 years ago
14

joanna split three pitchers of water equally among her eight plants.what fraction of a pitcher did each plant get?

Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
6 0
3/8 of a fraction is the amount of water that was split 

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Question 6 options: What is the approximate area of a circle with a diameter of 56 cm? Use your calculator button for π. Round y
Andrew [12]

The area of circle is 2463.0086 cm².

Step-by-step explanation:

Given,

Diameter of circle = 56 cm

Radius of circle = \frac{Diameter}{2}

Radius of circle = \frac{56}{2}=28\ cm

We know that;

Area of circle = \pi r^2

Area of circle = \pi *(28)^2

Area of circle = π * 784

Area of circle = 2463.00864041 cm²

Rounding off to four decimal places

Area of circle = 2463.0086 cm²

The area of circle is 2463.0086 cm².

Keywords: area, circle

Learn more about area at:

  • brainly.com/question/11416224
  • brainly.com/question/11638377

#LearnwithBrainly

8 0
3 years ago
School taxes for a local district are 2.5% of a household's income. The Smith household earns $80,000 per year. How much do the
Alex
Since 2.5% as a decimal is .025, the equation to get this is:

(.025)(80000)

Once multiplied, these equal 2,000. Or in other words, the Smiths pay $2,000 in school taxes per year.
5 0
2 years ago
The desired percentage of SiO2 in a certain type of aluminous cement is 5.5. To test whether the true average percentage is 5.5
miskamm [114]

Answer:  

z=-3.25

p_v =2*P(z  

Step-by-step explanation:  

1) Data given and notation  

\bar X=5.24 represent the mean production for the sample  

\sigma=0.32 represent the sample standard deviation for the sample  

n=16 sample size  

\mu_o =5.5 represent the value that we want to test  

\alph=0.05a represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

Part a: State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean production is different from 5.5 tons, the system of hypothesis would be:  

Null hypothesis:\mu =5.5  

Alternative hypothesis:\mu \neq 5.5  

If we analyze the info given we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

z=\frac{5.24-5.5}{\frac{0.32}{\sqrt{16}}}=-3.25  

P-value

Since this is a two side test the p value would be:  

p_v =2*P(z  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the production its significant different compared to the desired percentage of SiO2 of 5.5 at 5% of signficance.  

7 0
2 years ago
= Solving a word problem with two unknowns using a linear... A washer and a dryer cost $650 combined. The washer costs $100 less
nikdorinn [45]

Answer:

Washer 225

Dryer 425

.....

7 0
2 years ago
a building with a height of 32m casts a shadow that is 20m long. a person standing casts a shadow that is 1.2 m long how tall is
wlad13 [49]
Most of the information's required for solving the question is already given in the question.
Height of the building that casts a shadow of 20 m = 32 m
Then
Height of the man that casts a shadow of 1.2 m = (32/20) * 1.2 meter
                                                                             = 3.2 * 1.2 meter
                                                                             = 3.84 meter
So the actual height of the person casting a shadow of 1.2 meter is 3.84 meters. I hope that the procedure used for solving the problem is easy enough for you to understand. You can definitely use this method in future for solving problems of similar type without requiring any additional help from outside.

6 0
3 years ago
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