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3 years ago

I can't figure this one out...

Mathematics

2 answers:

lubasha [3.4K]3 years ago

7 0

Answer:

Exact: Area 36π ft^2 and Circumference 12π ft.

Approximate; Area = 113.10 ft^2 and Circumference 37.70 ft (to nearest hundredth).

Step-by-step explanation:

The diameter is 12 ft so the radius = 1/2 * 12 = 6 ft.

Area = π r^2

= 6^2 π = 36π ft^2. (exact).

Circumference = 2π r

= 12π ft (exact).

The approximate values are Area 113.10 and Circumference 37.70 to nearest hundredth.

worty [1.4K]3 years ago

5 0

Answer:

the area would be 36(pi) since you can't find out the exact value of pi (22/7) and where it would terminate so thats the most exact answer you can get. The circumference would be

Step-by-step explanation:

12/2 = 6

Area = (pi)radius^2

6^2 = 36

36(pi)

Circumference: (pi)d

12(pi)

For approximate area you multiply r^2 by 3.14

For approximate circumference you multiply diameter by 3.14

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Answer:

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3 years ago

John's parents purchased their first home in the 1980s with a 30-year

slega [8]

Answer:

Option A,$ 1,832.91 is correct

Step-by-step explanation:

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fv is the total amount of repayment which is unknown

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3 years ago

(15 points) Which expression has a value of 5 x 10^-3

velikii [3]

Answer:

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Step-by-step explanation:

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2 years ago

Simplify (1 − cos x)(1 + cos x). (2 points)

Vadim26 [7]

Answer:

sin²x

Step-by-step explanation:

we have

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4 0

3 years ago

Read 2 more answers

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

I am Lyosha [343]

Answer:

(a) The proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b) The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

Step-by-step explanation:

Let X = number of students who read above the eighth grade level.

(a)

A sample of n = 269 students are selected. Of these 269 students, X = 224 students who can read above the eighth grade level.

Compute the proportion of students who can read above the eighth grade level as follows:

$\hat p=\frac{X}{n}=\frac{224}{269}=0.8327$

The proportion of students who can read above the eighth grade level is 0.8327.

Compute the proportion of tenth graders reading at or below the eighth grade level as follows:

$1-\hat p=1-0.8327$

$=0.1673$

Thus, the proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b)

the information provided is:

n = 709

X = 546

Compute the sample proportion of tenth graders reading at or below the eighth grade level as follows:

$\hat q=1-\hat p$

$=1-\frac{X}{n}$

$=1-\frac{546}{709}$

$=0.2299\\\approx 0.229$

The critical value of z for 95% confidence interval is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the 95% confidence interval for the population proportion as follows:

$CI=\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

$=0.229\pm 1.96\times \sqrt{\frac{0.229(1-0.229)}{709}}\\=0.229\pm 0.03136\\=(0.19764, 0.26036)\\\approx (0.198, 0.260)$

Thus, the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

5 0

3 years ago