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3 years ago

12. Given a right triangle with the length

Mathematics

1 answer:

faltersainse [42]3 years ago

6 0

Answer:

G. 12 cm

Step-by-step explanation:

Length of one leg of right triangle = a = 9 cm

Length of hypotenuse = c = 15 cm

Length of the other leg = b = ?

Applying pythagorean theorem, c² = a² + b², we have:

15² = 9² + b²

15² - 9² = b²

144 = b²

b = √144

b = 12

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Kamila [148]

Answer:

Step-by-step explanation:

46-32 x 5/9 = 7.8

82-32x 5/9 = 27.8

8 0

4 years ago

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true

Feliz [49]

Answer:

a) The 95% CI for the true average porosity is (4.51, 5.19).

b) The 98% CI for true average porosity is (4.11, 5.01)

c) A sample size of 15 is needed.

d) A sample size of 101 is needed.

Step-by-step explanation:

a. Compute a 95% CI for the true average porosity of a certain seam if the average porosity for 20 specimens from the seam was 4.85.

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{0.78}{\sqrt{20}} = 0.34$

The lower end of the interval is the sample mean subtracted by M. So it is 4.85 - 0.34 = 4.51

The upper end of the interval is the sample mean added to M. So it is 4.35 + 0.34 = 5.19

The 95% CI for the true average porosity is (4.51, 5.19).

b. Compute a 98% CI for true average porosity of another seam based on 16 specimens with a sample average of 4.56.

Following the same logic as a.

98% C.I., so $z = 2.327$

$M = 2.327*\frac{0.78}{\sqrt{16}} = 0.45$

4.56 - 0.45 = 4.11

4.56 + 0.45 = 5.01

The 98% CI for true average porosity is (4.11, 5.01)

c. How large a sample size is necessary if the width of the 95% interval is to be 0.40?

A sample size of n is needed.

n is found when M = 0.4.

95% C.I., so Z = 1.96.

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.4 = 1.96*\frac{0.78}{\sqrt{n}}$

$0.4\sqrt{n} = 1.96*0.78$

$\sqrt{n} = \frac{1.96*0.78}{0.4}$

$(\sqrt{n})^{2} = (\frac{1.96*0.78}{0.4})^{2}$

$n = 14.6$

Rounding up

A sample size of 15 is needed.

d. What sample size is necessary to estimate the true average porosity to within 0.2 with 99% confidence?

99% C.I., so z = 2.575

n when M = 0.2.

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.2 = 2.575*\frac{0.78}{\sqrt{n}}$

$0.2\sqrt{n} = 2.575*0.78$

$\sqrt{n} = \frac{2.575*0.78}{0.2}$

$(\sqrt{n})^{2} = (\frac{2.575*0.78}{0.2})^{2}$

$n = 100.85$

Rounding up

A sample size of 101 is needed.

8 0

4 years ago

Find the first four terms of the recursive sequence defined by the following formula:

NISA [10]

Answer:

144, 36, 9, 2 $\frac{1}{4}$

Step-by-step explanation:

The recursive formula allows us to find a term in a sequence from the previous term.

Given

$a_{n}$ = $\frac{a_{n-1} }{4}$

Given the fourth term we require to work back to the third term , second and so on. Rearrange the formula to give

Multiply both sides by 4, then

$a_{n-1}$ = 4 $a_{n}$

Given a₄ = 2 $\frac{1}{4}$ , then

a₃ = 4 × a₄ = 4 × 2 $\frac{1}{4}$ = 9

a₂ = 4 × a₃ = 4 × 9 = 36

a₁ = 4 × a₂ = 4 × 36 = 144

4 0

4 years ago

Read 2 more answers

Express this folwing raito in the simplest form 12:52

Sunny_sXe [5.5K]

Answer:

3:13

Step-by-step explanation:

hope this helps you and good luck

5 0

3 years ago

Read 2 more answers

36 - 40. In order to save for her high school graduation, Marie decided to save P200 pesos at the end of each month. If the

PtichkaEL [24]

Answer:

200.4 at 0.25%

Step-by-step explanation:

Given data

P= P200

r= 0.25%

t= 1 year

n= 12

A= P(1+ r/n)^nt

substitute

A= 200(1+ 0.0025/12)^12*1

A= 200(1+ 0.00020833333)^12

A= 200(1.0002)^12

A= 200* 1.002

A= 200.4

Hence the amount is 200.4 at 0.25%

7 0

3 years ago