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Nimfa-mama [501]
3 years ago
5

A local park has a grassy playground that is 3/5 mile long and 1/3 mile wide. What part of a square mile does the playground cov

er? Enter your answer, in simplest form, in the boxes.
?

----mi²

?
Mathematics
1 answer:
adell [148]3 years ago
5 0

Answer:

The playground covers \frac{1}{5} of a square mile


Step-by-step explanation:

The playground is a rectangular ground. The area of a rectangle is given by legth * width, hence,

The area of the playground is length * width

Area = \frac{3}{5}*\frac{1}{3}=\frac{3}{15}=\frac{1}{5} sq. mi


Hence the playground cover \frac{1}{5} of a square mile

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Find a picture that has a zero sloped line in it. Can’t be a picture of a graph.
lara [203]

Answer:

Step-by-step explanation:

Anything that is horizontal is a 0-slope. So take a pen for example if you put it on its side it is a 0-slope.

5 0
3 years ago
Consider a chemical company that wishes to determine whether a new catalyst, catalyst XA-100, changes the mean hourly yield of i
kolezko [41]

Answer:

Null hypothesis:\mu = 750  

Alternative hypothesis:\mu \neq 750  

t=\frac{811-750}{\frac{19.647}{\sqrt{5}}}=6.943  

p_v =2*P(t_{4}>6.943)=0.00226  

If we compare the p value and a significance level assumed \alpha=0.05 we see that p_v so we can conclude that we reject the null hypothesis, and the actual true mean is significantly different from 750 pounds per hour.  

Step-by-step explanation:

Data given and notation

Data:    801, 814, 784, 836,820

We can calculate the sample mean and sample deviation with the following formulas:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X=811 represent the sample mean  

s=19.647 represent the standard deviation for the sample

n=5 sample size  

\mu_o =750 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

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

State the null and alternative hypotheses to be tested  

We need to conduct a hypothesis in order to determine if the mean is different from 750 pounds per hour, the system of hypothesis would be:  

Null hypothesis:\mu = 750  

Alternative hypothesis:\mu \neq 750  

Compute the test statistic  

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

t-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".  

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

t=\frac{811-750}{\frac{19.647}{\sqrt{5}}}=6.943  

Now we need to find the degrees of freedom for the t distirbution given by:

df=n-1=5-1=4

What do you conclude?  

Compute the p-value  

Since is a two tailed test the p value would be:  

p_v =2*P(t_{4}>6.943)=0.00226  

If we compare the p value and a significance level assumed \alpha=0.05 we see that p_v so we can conclude that we reject the null hypothesis, and the actual true mean is significantly different from 750 pounds per hour.  

4 0
3 years ago
Ryann planned on swimming 14 miles. Instead, he was only able to swim 1/4 of that distance. How many miles did Ryan swim?
mote1985 [20]

Hey there,

Given the question it shows that Ryann swam 1/4 of 14 miles that he was supposed to swim.

In words of is another way to say 1/4 * 14...

  • 14/4

This expressed as a mixed number would be...

  • 3 and 2/4 or 3 1/2

Of the 14 miles that Ryann was supposed to swim he swam 3 1/2 miles.

Hope I helped,

Amna

8 0
3 years ago
Read 2 more answers
Solve by substitution y=2x-1 y=3x+1
kupik [55]
Answer: x = -2

Step-by-Step Explanation:

Since the two expressions both equal y, you can set them equal to each other.

2x-1 = 3x+1

Subtract 2x from both sides.

-1 = x+1

Subtract 1 from both sides.

-2 = x

3 0
3 years ago
A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples 41 specimens and counts the number of
e-lub [12.9K]

Answer:

(53.3; 56.1)

Step-by-step explanation:

Given that:

Sample size, n = 41

Mean, xbar = 54.7

Standard deviation, s = 5.3

Confidence level, Zcritical at 90% = 1.645

Confidence interval :

Xbar ± Margin of error

Margin of Error = Zcritical * s/sqrt(n)

Margin of Error = 1.645 * 5.3/sqrt(41)

Margin of Error = 1.362

Lower boundary = 54.7 - 1.362 = 53.338

Upper boundary = 54.7 + 1.362 = 56.062

(53.3 ; 56.1)

6 0
3 years ago
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