HELP PLEASE! I am so bad at this I feel so dumb

A 100 cm² b 9 cm² c 225 cm² d 28.26 cm²

nasty-shy [4]

Answer:

28.26 cm^2

Step-by-step explanation:

A = (pi)r^2

A = 3.14 * (3 cm)^2

A = 3.14 * 9 cm^2

A = 28.26 cm^2

6 0

3 years ago

-9=-3(x+2) solve for x show steps

lianna [129]

SimplifyOkie dokie so first things first, you have to get rid of those parentheses! To do this you use the distributive property. -3(x+2) is the same thing as -3(x)+-3(2). that and you get -9=-3x-6. So you want to get x by itself to be able to find what x is, so you add 6 to both sides of the equation to get rid of the -6. So then you get -3=-3x. Then you must divide both sides by the coefficient of x, which would be -3. When you divide both sides by -3 you get -3/-3=x, which is the same thing as 1=x. Hope this helped!

4 0

3 years ago

Read 2 more answers

one light flashes every 2 minutes and another flashes every 7 minutes. If both lights flash together at 1pm, what is the first t

vovangra [49]

Light A flashes every 2 minutes while light B flashes every 7. The goal is to figure out when both lights flash at the same time and then figure out the other part later. You have to find the common multiples of 7 and 2. Meaning that you have to find out which number can they both be multiplied into. In this case, it is 14 because 7 can be multiplied by 2 to get 14 and 2 can be multiplied by 7. So if every 14 minutes they flash together at the same time, now you need to find out what time AFTER 3 will they both flash. You know that they both flashed at 1:00 so they will flash again at 1:14 and 1:28 and so on, but you need to find out when is the soonest they will flash after 3. 2 hours after the 1:00 flash will be 3:00 so that is 120 minutes. 120 minutes divided by 14 minutes comes to 8.5 but you dont need the decimal. Just simply multiply 14 by 8 and that comes to 112 minutes and if you add that to 1:00 you get 2:52. so they flashed at 2:52 so the next time they flash will be 14 minutes after 2:52. Sorry for the super long answer

6 0

3 years ago

5. Artie uses 1 1/3 yards of rope to make the bottom of his hammock stronger He uses 10 inches of rope to strengthen some areas

LenKa [72]

The correct answer is C 11/8 and 48 inches

7 0

3 years ago

Assume that females have pulse rates that are normally distributed with a mean of μ=73.0 beats per minute and a standard deviati

Gennadij [26K]

Answer:

a. the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly selected, the probability that they have pulse rates with a mean less than 76 beats per minute is 0.8849

c. D. Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

Given that:

Mean μ =73.0

Standard deviation σ =12.5

a. If 1 adult female is randomly selected, find the probability that her pulse rate is less than 76 beats per minute.

Let X represent the random variable that is normally distributed with a mean of 73.0 beats per minute and a standard deviation of 12.5 beats per minute.

Then : X $\sim$ N ( μ = 73.0 , σ = 12.5)

The probability that her pulse rate is less than 76 beats per minute can be computed as:

$P(X < 76) = P(\dfrac{X-\mu}{\sigma}< \dfrac{X-\mu}{\sigma})$

$P(X < 76) = P(\dfrac{76-\mu}{\sigma}< \dfrac{76-73}{12.5})$

$P(X < 76) = P(Z< \dfrac{3}{12.5})$

$P(X < 76) = P(Z< 0.24)$

From the standard normal distribution tables,

$P(X < 76) = 0.5948$

Therefore , the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly selected, find the probability that they have pulse rates with a mean less than 76 beats per minute.

now; we have a sample size n = 25

The probability can now be calculated as follows:

$P(\overline X < 76) = P(\dfrac{\overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{ \overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}})$

$P( \overline X < 76) = P(\dfrac{76-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{76-73}{\dfrac{12.5}{\sqrt{25}}})$

$P( \overline X < 76) = P(Z< \dfrac{3}{\dfrac{12.5}{5}})$

$P( \overline X < 76) = P(Z< 1.2)$

From the standard normal distribution tables,

$P(\overline X < 76) = 0.8849$

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

In order to determine the probability in part (b); the normal distribution is perfect to be used here even when the sample size does not exceed 30.

Therefore option D is correct.

Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

5 0

3 years ago