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1 year ago

I need help with all A B C, I’m going to type the C QUESTION BECAUSE IT DIDNT APPER ON THE PIC

Mathematics

1 answer:

igomit [66]1 year ago

8 0

1. Class width:

The class width is given by the following expression:

$Width=\frac{Range}{#intervals}$

The number of intervals are: 4

And the range is:

$Range=maximumV-MinimuV=509-210=299$

Replacing:

$Width=\frac{299}{4}=74.75\approx74$

A) The class width is 74 (rounding to the smallest value).

2. Number of team attempted from 360 to 434:

B) The number of teams are: 4

3. According to the histogram, what is the total number of teams?

The total number of teams is given by:

$Total=8+3+4+8=23$

C) 23.

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Determine if the series is convergent or divergent 20-15+10-5....

Kazeer [188]

Answer:

Option b is correct.

The series 20-15+10-5....is Divergent

Step-by-step explanation:

Alternating series Test:

$\sum_{n=1}^{\infty} (-1)^{n-1} (b_n) =b_1-b_2+b_3-........$

$b_n>0$ satisfies:

$b_{n+1} \leq b_n$ for all n

$\lim_{n\rightarrow \infty} b_n = 0$

Then the series converges,

otherwise diverges.

Given the series: 20-15+10-5....

This is a alternating series:

$\sum_{n=1}^{\infty} (-1)^{n-1} (20-5(n-1))$

$b_n = (20-5(n-1))$

$b_{n+1} = (20-5(n+1-1)) = (20-5n)$

using the alternating series test;

$b_{n+1} \leq b_n$ for all n

$\lim_{n\rightarrow \infty} b_n = \lim_{n\rightarrow \infty} (20-5(n-1)) = -\infty$

⇒ the series diverges.

therefore, the given series i,e 20-15+10-5.... is divergent.

3 0

4 years ago

Add to negative 20 and multiply to 21

likoan [24]

-20+13=-7 , -7 multiplied by -3 equals 21

7 0

3 years ago

Need help desperately.. y – 2.234 = 7.92

mart [117]

Answer:

251,734.

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

A basketball player scored x, y, and z points in 3 games. Express the average number of points scored by the player in terms of

Anettt [7]

The average will be the total points divided by the number of games:

(x+y+z) / 3

7 0

4 years ago

The computers of six faculty members in a certain department are to be replaced. Two of the faculty members have selected laptop

Anettt [7]

Answer:

a. $\frac{1}{15}$

b. $\frac{2}{5}$

c. $\frac{14}{15}$

d. $\frac{8}{15}$

Step-by-step explanation:

Given that there are two laptop machines and four desktop machines.

On a day, 2 computers to be set up.

To find:

a. probability that both selected setups are for laptop computers?

b. probability that both selected setups are desktop machines?

c. probability that at least one selected setup is for a desktop computer?

d. probability that at least one computer of each type is chosen for setup?

Solution:

Formula for probability of an event E can be observed as:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

a. Favorable cases for Both the laptops to be selected = $_2C_2$ = 1

Total number of cases = 15

Required probability is $\frac{1}{15}$ .

b. Favorable cases for both the desktop machines selected = $_4C_2=6$

Total number of cases = 15

Required probability is $\frac{6}{15} = \frac{2}{5}$ .

c. At least one desktop:

Two cases:

1. 1 desktop and 1 laptop:

Favorable cases = $_2C_1\times _4C_1 = 8$

2. Both desktop:

Favorable cases = $_4C_2=6$

Total number of favorable cases = 8 + 6 = 14

Required probability is $\frac{14}{15}$ .

d. 1 desktop and 1 laptop:

Favorable cases = $_2C_1\times _4C_1 = 8$

Total number of cases = 15

Required probability is $\frac{8}{15}$ .

8 0

4 years ago