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dimulka [17.4K]

3 years ago

10

Kendra has a painting canvas that is 24 inches wide by 38 inches high. She painted a red rose, which covered 30% of the canvas.

What was the area of the canvas which was covered by the red rose?

1 answer:

V125BC [204]3 years ago

7 0

Answer:

x=370 in ∧ 2

Step-by-step explanation:

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Blood pressure values are often reported to the nearest 5 mmHg (100, 105, 110, etc.). The actual blood pressure values for nine

Paladinen [302]

Answer:

a) 117.4

b) 117.9

c) Option A) When there is rounding or grouping, the median can be highly sensitive to small change

Step-by-step explanation:

We are given the following data set in the question:

108.6, 117.4, 128.4, 120.0, 103.7, 112.0, 98.3, 121.5, 123.2

$Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}$

n = 9

a) Median of the reported blood pressure values

Sorted Values: 98.3, 103.7, 108.6, 112.0, 117.4, 120.0, 121.5, 123.2, 128.4

Median =

$\dfrac{9 + 1}{2}^{th}\text{ term} = 5^{th}\text{ term} = 117.4$

b) New median of the reported values

Data: 108.6, 117.9, 128.4, 120.0, 103.7, 112.0, 98.3, 121.5, 123.2

Sorted Values: 98.3, 103.7, 108.6, 112.0, 117.9, 120.0, 121.5, 123.2, 128.4

New Median =

$\dfrac{9 + 1}{2}^{th}\text{ term} = 5^{th}\text{ term} = 117.9$

c) Since median is a position based descriptive statistics, a small change in values can bring a change in the median value as the order of the data may change.

Option A) When there is rounding or grouping, the median can be highly sensitive to small change

7 0

3 years ago

What comes after 5,10,20,40,80...

fgiga [73]

It is just doubled the number before so the next number would be 160, then it would be 360, 720, etc..

7 0

3 years ago

Ellie has 9 weeks to finish an 11 part project. How many parts does Ellie have to do each week?

netineya [11]

11 divided by 9 is 1.2 repeating so Ellie will have to do about 1.3 parts a week

5 0

3 years ago

A city that had 7500 thousand people at the beginning of the year 2000 has been decreasing by 3.6% per year. a. What is the 1-ye

storchak [24]

Answer:

Step-by-step explanation:

a) if the population at the beginning of the year 2000 was 7500 people,

The 1-year percent change in the city's population would be

3.6/100 × 7500 = 270

b) The population after 1 year is

7500 - 270 = 7230

The percentage of the previous value of the population to its new value for each year is

7230/7500 × 100 = 96.4%

c) the 1-year growth factor for the population of the city would be

(1 - 0.036)^1 = 0.964

d) the function, g that determines the population of the city (in thousands of people) in terms of the number of years t since the beginning of 2000 would be

g = 7500(1 - 0.036)^t

g = 7500(0.964)^t

5 0

3 years ago

We are conducting a hypothesis test to determine if fewer than 80% of ST 311 Hybrid students complete all modules before class.

Juli2301 [7.4K]

Answer:

$z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124$

Null hypothesis: $p\geq 0.8$

Alternative hypothesis: $p < 0.8$

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

$p_v =P(Z$

So the p value obtained was a very high value and using the significance level assumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Be Careful with the system of hypothesis!

If we conduct the test with the following hypothesis:

Null hypothesis: $p\leq 0.8$

Alternative hypothesis: $p > 0.8$

$p_v =P(Z>2.124)=0.013$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation

n=110 represent the random sample taken

X=97 represent the students who completed the modules before class

$\hat p=\frac{97}{110}=0.882$ estimated proportion of students who completed the modules before class

$p_o=0.8$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v{/tex} represent the p value (variable of interest) 2) Concepts and formulas to use We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.8 or 80%: Null hypothesis:[tex]p\geq 0.8$

Alternative hypothesis: $p < 0.8$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124$

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level assumed $\alpha=0.05$ . The next step would be calculate the p value for this test.

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

$p_v =P(Z$

So the p value obtained was a very high value and using the significance level assumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Be Careful with the system of hypothesis!

If we conduct the test with the following hypothesis:

Null hypothesis: $p\leq 0.8$

Alternative hypothesis: $p > 0.8$

$p_v =P(Z>2.124)=0.013$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

5 0

4 years ago