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Radda [10]
2 years ago
8

Bro i need help with this 8th grade math work

Mathematics
1 answer:
Zolol [24]2 years ago
8 0

Answer:

look it up

Step-by-step explanation:

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The irrigation arm is 300 ft. 300 ft is just the radius. The diameter is 600, because to get the diameter you must multiply the radius by two. So 600 is the land the irrigation arm covers.
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3 years ago
The diameter of the moon is 2,160 miles. A model has a scale of 1 in : 150 mi. What is the diameter of the model?
Charra [1.4K]

Answer:

The diameter of the model is 14.4 inches.

Step-by-step explanation:

The Diameter of the moon = 2,160 miles

The scale on the model represents 1 in = 150 miles

Let the model represents k inches in 2,160 miles.

So, by the Ratio of Proportionality:

\frac{1}{150}   = \frac{k}{2160}

⇒k = \frac{2160}{150} \\\implies k =  14.4

or, k = 14.4 inches

⇒On the scale 2160 miles is represented as 14.4 inches

    Hence the diameter of the model is 14.4 inches.

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3 years ago
Sales tax in the state of Florida is 7%. Bob purchased a coat that was marked $350.00. What is the total bill including tax?
pav-90 [236]
To find the tax, you multiply the tax rate by the amount to be taxed, so...350 × 7%
350 × .07 = $24.50

This is the amount of tax. Now add it to the original price for the total cost.

$350.00 + $24.50 =

$374.50
6 0
4 years ago
How many hours is 80 minutes?
Oduvanchick [21]
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7 0
3 years ago
Read 2 more answers
Compute the standard error for sample proportions from a population with proportion p= 0.55 for sample sizes of n=30, n=100 and
Pani-rosa [81]

Given Information:

Population proportion = p =  0.55

Sample size 1 = n₁ = 30

Sample size 2 = n₂ = 100

Sample size 3 = n₃ = 1000

Required Information:

Standard error = σ = ?

Answer:

$ \sigma_1 = 0.091 $

$ \sigma_2 = 0.050 $

$ \sigma_3 = 0.016 $

Step-by-step explanation:

The standard error for sample proportions from a population is given by

$ \sigma =  \sqrt{\frac{p(1-p)}{n} } $  

Where p is the population proportion and n is the sample size.

For sample size n₁ = 30

$ \sigma_1 =  \sqrt{\frac{p(1-p)}{n_1} } $

$ \sigma_1 =  \sqrt{\frac{0.55(1-0.55)}{30} } $

$ \sigma_1 = 0.091 $

For sample size n₂ = 100

$ \sigma_2 =  \sqrt{\frac{p(1-p)}{n_2} } $

$ \sigma_2 =  \sqrt{\frac{0.55(1-0.55)}{100} } $

$ \sigma_2 = 0.050 $

For sample size n₃ = 1000

$ \sigma_3 =  \sqrt{\frac{p(1-p)}{n_3} } $

$ \sigma_3 =  \sqrt{\frac{0.55(1-0.55)}{1000} } $

$ \sigma_3 = 0.016 $

As you can notice, the standard error decreases as the sample size increases.

Therefore, the greater the sample size lesser will be the standard error.

8 0
4 years ago
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