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IceJOKER [234]
3 years ago
9

Eric drew a scale drawing of a country park. The scale he used was 1 inch = 2.5 yards. The picnic area is 80 yards wide in real

life. How wide is the picnic area in the drawing ?
Mathematics
1 answer:
Alchen [17]3 years ago
5 0
If 1 inch = 2.5 yards then
x inch = 80 yards
cross multiplying we have:
80 × 1 = x × 2.5
x = 80÷ 2.5 = 32 inches
therefore the picnic area is 32 inches in the drawing.
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Write an equation of the line that passes through (18, 2) and is parallel to the line 3y - x = -12.
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Answer:

3y=x-12

y=x/3-4

m=1/3

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y-2=1/3(x-18)

y-2=x/3 - 6

y=x/3 - 4

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3 years ago
Solve inequality for x -4(2x+4)&gt;8x+64
Mkey [24]

Answer:

x< - \frac{16}{3}

Step-by-step explanation:

x 8x_16>8x+64

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4 0
1 year ago
Read 2 more answers
Andell is 9 years older than Arnez. arnez is 3 years younger than Marianne. the sum of their ages is 24. what is the age of each
Harman [31]

Answer:

Arnez is 4. Andell is 13. Marianne is 7.

Step-by-step explanation:

Let x represent Arnez's age. If we write out an equation, we end up with this:

24 = (9 + x) + (x + 3) + x

24 = 9 + x + x + 3 + x

24 = (x + x + x) + (9 + 3)

24 = 3x + 12

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5 0
3 years ago
You work in the HR department at a large franchise. you want to test whether you have set your employee monthly allowances corre
ra1l [238]

Answer:

1) Null hypothesis:\mu \leq 500  

Alternative hypothesis:\mu > 500  

z=\frac{640-500}{\frac{150}{\sqrt{40}}}=5.90  

For this case since we are conducting a right tailed test we need to find a critical value in the normal standard distribution who accumulates 0.01 of the area in the right and we got:

z_{crit}= 2.33

For this case we see that the calculated value is higher than the critical value

Since the calculated value is higher than the critical value we have enugh evidence to reject the null hypothesis at 1% of significance level

2) Since is a right tailed test the p value would be:  

p_v =P(z>5.90)=1.82x10^{-9}  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, same conclusion for part 1

Step-by-step explanation:

Part 1

Data given

\bar X=640 represent the sample mean

\sigma=150 represent the population standard deviation

n=40 sample size  

\mu_o =500 represent the value that we want to test  

\alpha=0.01 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)  

Step1:State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is higher than 500, the system of hypothesis would be:  

Null hypothesis:\mu \leq 500  

Alternative hypothesis:\mu > 500  

Step 2: Calculate the statistic

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

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

z=\frac{640-500}{\frac{150}{\sqrt{40}}}=5.90  

Step 3: Calculate the critical value

For this case since we are conducting a right tailed test we need to find a critical value in the normal standard distribution who accumulates 0.01 of the area in the right and we got:

z_{crit}= 2.33

Step 4: Compare the statistic with the critical value

For this case we see that the calculated value is higher than the critical value

Step 5: Decision

Since the calculated value is higher than the critical value we have enugh evidence to reject the null hypothesis at 1% of significance level

Part 2

P-value  

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

p_v =P(z>5.90)=1.82x10^{-9}  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, same conclusion for part 1

7 0
3 years ago
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