Can someone please help me with this of identifying the domain and range!!!!

gulaghasi [49]

C. Alternate interior angles are formed by a transverse line intersecting two parallel lines . They are located between the two parallel lines but on opposite sides of the transverse line

5 0

2 years ago

Read 2 more answers

The overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 7.8 cm.

devlian [24]

Answer:

Given the mean = 205 cm and standard deviation as 7.8cm

a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z $\leq 1.72$ ). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.

b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15) from 1. Therefore, we have 1- P(Z $\leq -1.09$ ). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.

c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.

Step-by-step explanation:

Given the mean = 205 cm and standard deviation as 7.8cm

a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z $\leq 1.72$ ). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.

b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15) from 1. Therefore, we have 1- P(Z $\leq -1.09$ ). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.

c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.

4 0

3 years ago

Can someone help me

marshall27 [118]

Answer:

16682.7 cm³

Step-by-step explanation:

The total volume is the volume of the cone plus half the volume of a sphere:

V = ⅓ π r² h + ½ (4/3 π r³)

V = ⅓ π r² h + ⅔ π r³

V = ⅓ π r² (h + 2r)

Given that r = 4.15 cm and h = 10.2 cm:

V = ⅓ π (4.15)² (10.2 + 2×4.15)

V ≈ 333.65

The volume needed for 50 cones is therefore:

50V ≈ 16682.7 cm³

7 0

3 years ago

Use the initial term and the recursive formula to find an explicit formula for the sequence an. Write your answer in simplest fo

agasfer [191]

From the recursive rule

$a_n = a_{n-1}-13$

it follows that

$a_{n-1}=a_{n-2}-13 \implies a_n = a_{n-2} - 2\times13$

$a_{n-2}=a_{n-3}-13 \implies a_n = a_{n-3} - 3\times13$

and so on. Notice how the subscript on a on the right side and the coefficient multiplied by 13 add up to n (n - 2 + 2 = n; n - 3 + 3 = n; and so on). If we continue the pattern, we'll end up with

$a_n = a_1 + (n-1)\times13$

so that the explicit rule for the n-th term in the sequence is

$a_n = -2 + 13(n-1) \implies a_n = \boxed{13n-15}$

6 0

2 years ago

Help please solve<br> <img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B6x%5E5%2B11x%5E4-11x-6%7D%7B%282x%5E2-3x%2B1

Shkiper50 [21]

Answer:

$\displaystyle -\frac{1}{2} \leq x < 1$

Step-by-step explanation:

<u>Inequalities</u>

They relate one or more variables with comparison operators other than the equality.

We must find the set of values for x that make the expression stand

$\displaystyle \frac{6x^5+11x^4-11x-6}{(2x^2-3x+1)^2} \leq 0$

The roots of numerator can be found by trial and error. The only real roots are x=1 and x=-1/2.

The roots of the denominator are easy to find since it's a second-degree polynomial: x=1, x=1/2. Hence, the given expression can be factored as

$\displaystyle \frac{(x-1)(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)^2(x-\frac{1}{2})^2} \leq 0$

Simplifying by x-1 and taking x=1 out of the possible solutions:

$\displaystyle \frac{(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)(x-\frac{1}{2})^2} \leq 0$

We need to find the values of x that make the expression less or equal to 0, i.e. negative or zero. The expressions

$(6x^3+14x^2+10x+12)$

is always positive and doesn't affect the result. It can be neglected. The expression

$(x-\frac{1}{2})^2$

can be 0 or positive. We exclude the value x=1/2 from the solution and neglect the expression as being always positive. This leads to analyze the remaining expression

$\displaystyle \frac{(x+\frac{1}{2})}{(x-1)} \leq 0$